Research ArticleSharp Bounds for Neuman Means by HarmonicArithmetic and Contraharmonic Means

Zhi-Jun Guo1 Yu-Ming Chu1 Ying-Qing Song1 and Xiao-Jing Tao2

1 School of Mathematics and Computation Sciences Hunan City University Yiyang 413000 China2 College of Mathematics and Econometrics Hunan University Changsha 410082 China

Correspondence should be addressed to Yu-Ming Chu chuyuming2005126com and Ying-Qing Song 1452225875qqcom

Received 20 May 2014 Accepted 12 July 2014 Published 23 July 2014

Academic Editor Shuangjie Peng

Copyright copy 2014 Zhi-Jun Guo et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We give several sharp bounds for the Neumanmeans119873119860119867

and119873119867119860

(119873119862119860

and119873119860119862) in terms of harmonicmeanH (contraharmonic

mean C) or the geometric convex combination of arithmetic mean A and harmonic mean H (contraharmonic mean C andarithmetic mean A) and present a new chain of inequalities for certain bivariate means

1 Introduction

For 119886 119887 gt 0 with 119886 = 119887 the Schwab-Borchardt mean SB(119886 119887)[1ndash3] of 119886 and 119887 is defined as

SB (119886 119887) =

radic1198872 minus 1198862

cosminus1 (119886119887) 119886 lt 119887

radic1198862 minus 1198872

coshminus1 (119886119887) 119886 gt 119887

(1)

where cosminus1(119909) and coshminus1(119909) = log(119909 + radic1199092 minus 1) arethe inverse cosine and inverse hyperbolic cosine functionsrespectively

The Schwab-Borchardt mean SB(119886 119887) can be expressedby the symmetric elliptic integral 119877

119865[4] of the first kind as

follows [5] (see also [6 (321)])

SB (119886 119887) = 1

119877119865(1198862 1198872 1198872)

(2)

where 119877119865(119886 119887 119888) = (12) int

infin

0[(119905 + 119886)(119905 + 119887)(119905 + 119888)]

minus12119889119905

Recently the Schwab-Borchardt mean has been the sub-ject of intensive research In particular many remarkableinequalities for the Schwab-Borchardtmean and its generatedmeans can be found in the literature [1ndash3 7ndash10]

Very recently Neuman [11] found a new mean 119873(119886 119887)

derived from the Schwab-Borchardt mean as follows

119873(119886 119887) =1

2[119886 +

1198872

SB (119886 119887)] (3)

Let 119866(119886 119887) = radic119886119887 119867(119886 119887) = 2119886119887(119886 + 119887) 119871(119886 119887) =

(119886 minus 119887)(log 119886 minus log 119887) 119875(119886 119887) = (119886 minus 119887)[2sinminus1((119886 minus 119887)(119886 +119887))] 119860(119886 119887) = (119886 + 119887)2 119872(119886 119887) = (119886 minus 119887)[2sinhminus1((119886 minus119887)(119886 + 119887))] 119879(119886 119887) = (119886 minus 119887)[2tanminus1((119886 minus 119887)(119886 + 119887))]119876(119886 119887) = radic(1198862 + 1198872)2 and 119862(119886 119887) = (119886

2+ 1198872)(119886 +

119887) be respectively the geometric harmonic logarithmicfirst Seiffert arithmetic Neuman-Sandor second Seiffertquadratic and contraharmonic means and let

119873119860119867 (119886 119887) = 119873 [119860 (119886 119887) 119867 (119886 119887)]

119873119867119860 (119886 119887) = 119873 [119867 (119886 119887) 119860 (119886 119887)]

119873119860119866 (119886 119887) = 119873 [119860 (119886 119887) 119866 (119886 119887)]

119873119866119860 (119886 119887) = 119873 [119866 (119886 119887) 119860 (119886 119887)]

119873119860119862 (119886 119887) = 119873 [119860 (119886 119887) 119862 (119886 119887)]

119873119862119860 (119886 119887) = 119873 [119862 (119886 119887) 119860 (119886 119887)]

119873119860119876 (119886 119887) = 119873 [119860 (119886 119887) 119876 (119886 119887)]

119873119876119860 (119886 119887) = 119873 [119876 (119886 119887) 119860 (119886 119887)]

(4)

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 914242 8 pageshttpdxdoiorg1011552014914242

2 Abstract and Applied Analysis

be the Neuman means Then Neuman [11] proved that

119866 (119886 119887) lt 119871 (119886 119887) lt 119873119860119866 (119886 119887) lt 119875 (119886 119887) lt 119873119866119860 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119873119876119860 (119886 119887) lt 119879 (119886 119887)

lt 119873119860119876 (119886 119887) lt 119876 (119886 119887) lt 119862 (119886 119887)

(5)

for all 119886 119887 gt 0 with 119886 = 119887 and the double inequalities

1205721119860 (119886 119887) + (1 minus 1205721) 119866 (119886 119887)

lt 119873119866119860 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731) 119866 (119886 119887)

1205722119876 (119886 119887) + (1 minus 1205722) 119860 (119886 119887)

lt 119873119860119876 (119886 119887) lt 1205732119876 (119886 119887) + (1 minus 1205732) 119860 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723) 119866 (119886 119887)

lt 119873119860119866 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733) 119866 (119886 119887)

1205724119876 (119886 119887) + (1 minus 1205724) 119860 (119886 119887)

lt 119873119876119860 (119886 119887) lt 1205734119876 (119886 119887) + (1 minus 1205734) 119860 (119886 119887)

(6)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 23 120573

1ge

1205874 1205722le 23 120573

2ge (120587 minus 2)[4(radic2 minus 1)] = 0689 120572

3le 13

1205733ge 12 120572

4le 13 and 120573

4ge [log(1 +radic2) +radic2 minus 2][2(radic2 minus

1)] = 0356 Zhang et al [12] presented the best possible parameters

1205721 1205722 1205731 1205732isin [0 12] and 120572

3 1205724 1205733 1205734isin [12 1] such that

the double inequalities

119866 (1205721119886 + (1 minus 120572

1) 119887 1205721119887 + (1 minus 120572

1) 119886)

lt 119873119860119866 (119886 119887) lt 119866 (1205731119886 + (1 minus 1205731) 119887 1205731119887 + (1 minus 1205731) 119886)

119866 (1205722119886 + (1 minus 120572

2) 119887 1205722119887 + (1 minus 120572

2) 119886)

lt 119873119866119860 (119886 119887) lt 119866 (1205732119886 + (1 minus 1205732) 119887 1205732119887 + (1 minus 1205732) 119886)

119876 (1205723119886 + (1 minus 120572

3) 119887 1205723119887 + (1 minus 120572

3) 119886)

lt 119873119876119860 (119886 119887) lt 119876 (120573

3119886 + (1 minus 120573

3) 119887 1205733119887 + (1 minus 120573

3) 119886)

119876 (1205724119886 + (1 minus 120572

4) 119887 1205724119887 + (1 minus 120572

4) 119886)

lt 119873119860119876 (119886 119887) lt 119876 (120573

4119886 + (1 minus 120573

4) 119887 1205734119887 + (1 minus 120573

4) 119886)

(7)

hold for all 119886 119887 gt 0 with 119886 = 119887In [13] the authors found the greatest values 120572

1 1205722 1205723

1205724 1205725 1205726 1205727 and 120572

8and the least values 120573

1 1205732 1205733 1205734 1205735 1205736

1205737 and 120573

8such that the double inequalities

1198601205721(119886 119887) 119866

1minus1205721(119886 119887)

lt 119873119866119860 (119886 119887) lt 119860

1205731(119886 119887) 119866

1minus1205731(119886 119887)

1205722

119866 (119886 119887)+

1 minus 1205722

119860 (119886 119887)

lt1

119873119866119860 (119886 119887)

lt1205732

119866 (119886 119887)+

1 minus 1205732

119860 (119886 119887)

1198601205723(119886 119887) 119866

1minus1205723(119886 119887)

lt 119873119860119866 (119886 119887) lt 119860

1205733(119886 119887) 119866

1minus1205733(119886 119887)

1205724

119866 (119886 119887)+

1 minus 1205724

119860 (119886 119887)

lt1

119873119860119866 (119886 119887)

lt1205734

119866 (119886 119887)+

1 minus 1205734

119860 (119886 119887)

1198761205725(119886 119887) 119860

1minus1205725(119886 119887)

lt 119873119860119876 (119886 119887) lt 119876

1205735(119886 119887) 119860

1minus1205735(119886 119887)

1205726

119860 (119886 119887)+

1 minus 1205726

119876 (119886 119887)

lt1

119873119860119876 (119886 119887)

lt1205736

119860 (119886 119887)+

1 minus 1205736

119876 (119886 119887)

1198761205727(119886 119887) 119860

1minus1205727(119886 119887)

lt 119873119876119860 (119886 119887) lt 119876

1205737(119886 119887) 119860

1minus1205737(119886 119887)

1205728

119860 (119886 119887)+

1 minus 1205728

119876 (119886 119887)

lt1

119873119876119860 (119886 119887)

lt1205738

119860 (119886 119887)+

1 minus 1205738

119876 (119886 119887)

(8)

hold for all 119886 119887 gt 0 with 119886 = 119887Let 119886 gt 119887 gt 0 V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)

119902 isin (0 1205872) 119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be theparameters such that 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) =sec(119904) = 1 + V2 Then He et al [14] proved that

119873119860119867 (119886 119887) =

1

2119860 (119886 119887) [1 +

2119901

sinh (2119901)] (9)

119873119867119860 (119886 119887) =

1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)] (10)

119873119862119860 (119886 119887) =

1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)] (11)

119873119860119862 (119886 119887) =

1

2119860 (119886 119887) [1 +

2119904

sin (2119904)] (12)

119867(119886 119887) lt 119873119860119867 (119886 119887) lt 119873119867119860 (119886 119887) lt 119860 (119886 119887)

lt 119873119862119860 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(13)

Let 119909 isin [0 12] 119910 isin [12 1]119891(119909) = 119867[119909119886+(1minus119909)119887 119909119887+(1 minus 119909)119886] and 119892(119910) = 119862[119910119886 + (1 minus 119910)119887 119910119887 + (1 minus 119910)119886] Thenwe clearly see that

Abstract and Applied Analysis 3

119891 (0) = 119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119873119867119860 (119886 119887) lt 119860 (119886 119887)

= 119891(1

2) = 119892(

1

2) lt 119873

119862119860 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

= 119892 (1)

(14)

for all 119886 119887 gt 0 with 119886 = 119887 and the functions 119891 and 119892 arerespectively strictly increasing on the intervals [0 12] and[12 1] for fixed 119886 119887 gt 0 with 119886 = 119887

Themain purpose of this paper is to find the best possibleparameters 120572

1 1205722 1205723 1205724 1205731 1205732 1205733 1205734isin R 120572

5 1205726 1205735 1205736isin

[0 12] and 1205727 1205728 1205737 1205738isin [12 1] such that the double

inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

119867 [1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119867 [120573

7119886 + (1 minus 120573

7) 119887 1205737119887 + (1 minus 120573

7) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(15)

hold for all 119886 119887 gt 0 with 119886 = 119887 and present a new chain ofinequalities for certain bivariate means

2 Lemmas

In order to prove our main results we need several lemmaswhich we present in this section

Lemma 1 (see [15 Theorem 125]) For minusinfin lt 119886 lt 119887 lt infinlet 119891 119892 [119886 119887] rarr R be continuous on [119886 119887] and differentiable

on (119886 119887) let 1198921015840(119909) = 0 on (119886 119887) If 1198911015840(119909)1198921015840(119909) is increasing(decreasing) on (119886 119887) then so are

119891 (119909) minus 119891 (119886)

119892 (119909) minus 119892 (119886)

119891 (119909) minus 119891 (119887)

119892 (119909) minus 119892 (119887) (16)

If 1198911015840(119909)1198921015840(119909) is strictly monotone then the monotonicity inthe conclusion is also strict

Lemma 2 (see [16 Lemma 11]) Suppose that the power series119891(119909) = sum

infin

119899=0119886119899119909119899 and 119892(119909) = sum

infin

119899=0119887119899119909119899 have the radius of

convergence 119903 gt 0 and 119886119899 119887119899gt 0 for all 119899 ge 0 If the sequence

119886119899119887119899 is (strictly) increasing (decreasing) for all 119899 ge 0 then

the function 119891(119909)119892(119909) is also (strictly) increasing (decreasing)on (0 119903)

Lemma 3 The function

1205931 (119909) =

sinh (119909) minus 119909sinh (119909) [cosh (119909) minus 1]

(17)

is strictly decreasing from (0infin) onto (0 13)

Proof Making use of power series expansion we get

1205931 (119909) =

2 sinh (119909) minus 2119909sinh (2119909) minus 2 sinh (119909)

=suminfin

119899=1(2 (2119899 + 1)) 119909

2119899+1

suminfin

119899=1((22119899+1 minus 2) (2119899 + 1)) 119909

2119899+1

=suminfin

119899=0(2 (2119899 + 3)) 119909

2119899

suminfin

119899=0((22119899+3 minus 2) (2119899 + 3)) 119909

2119899

(18)

Let

119886119899=

2

(2119899 + 3) 119887

119899=22119899+3

minus 2

(2119899 + 3) (19)

Then

119886119899gt 0 119887

119899gt 0 (20)

and 119886119899119887119899= 1(2

2119899+2minus 1) is strictly decreasing for all 119899 ge 0

Note that

1205931(0+) =

1198860

1198870

=1

3 120593

1 (infin) = lim119899rarrinfin

119886119899

119887119899

= 0 (21)

Therefore Lemma 3 follows easily from Lemma 2 and(18)ndash(21) together with the monotonicity of the sequence119886119899119887119899

Lemma 4 The function

1205932 (119909) =

119909 minus sin (119909)sin (119909) [1 minus cos (119909)]

(22)

is strictly increasing from (0 1205872) onto (13 1205872 minus 1)

4 Abstract and Applied Analysis

Proof Let 1198911(119909) = 119909 minus sin(119909) and 119892

1(119909) = sin(119909)[1 minus cos(119909)]

Then simple computations lead to

1205932 (119909) =

1198911 (119909)

1198921 (119909)

=1198911 (119909) minus 1198911 (0)

1198921 (119909) minus 1198921 (0)

(23)

and 11989110158401(119909)1198921015840

1(119909) = 1[1 + 2 cos(119909)] is strictly increasing on

(0 1205872)Note that

1205932(0+) = lim119909rarr0

+

1198911015840

1(119909)

1198921015840

1(119909)

=1

3 120593

2(120587

2) =

120587

2minus 1 (24)

Therefore Lemma 4 follows from Lemma 1 (23) (24)and the monotonicity of 1198911015840

1(119909)1198921015840

1(119909)

Lemma 5 The function

1205933 (119909) =

log [(119909 + sinh (119909) cosh (119909)) (2 sinh (119909) cosh (119909))]log (1 cosh (119909))

(25)

is strictly decreasing from (0infin) onto (0 23)

Proof Let 1198912(119909) = log[(119909 + sinh(119909) cosh(119909))(2 sinh(119909)

cosh(119909))] and 1198922(119909) = log[1 cosh(119909)] Then simple compu-

tations lead to

1205933 (119909) =

1198912 (119909)

1198922 (119909)

=1198912 (119909) minus 1198912 (0

+)

1198922 (119909) minus 1198922 (0)

(26)

1198911015840

2(119909)

1198921015840

2(119909)

= (119909 [cosh3 (119909) + sinh2 (119909) cosh (119909)] minus sinh (119909) cosh2 (119909))

times (sinh2 (119909) cosh (119909) [119909 + sinh (119909) cosh (119909)])minus1

= (8119909 cosh (3119909) + 8119909 cosh (119909) minus 4 sinh (3119909) minus 4 sinh (119909))

times (4119909 cosh (3119909) minus 4119909 cosh (119909) + sinh (5119909) minus sinh (3119909)

minus2 sinh(119909))minus1

= (

infin

sum

119899=1

( (8 (2119899 + 1) (1 + 32119899) minus 4 (1 + 3

2119899+1))

times ((2119899 + 1))minus1) 1199092119899+1

)

times (

infin

sum

119899=1

( (4 (2119899 + 1) (32119899minus 1) + 5

2119899+1minus 32119899+1

minus 2)

times ((2119899 + 1))minus1) 1199092119899+1

)

minus1

= (

infin

sum

119899=0

( (8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

))

times ((2119899 + 3))minus1) 1199092119899)

times (

infin

sum

119899=0

( (4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2)

times ((2119899 + 3))minus1) 1199092119899)

minus1

(27)

Let

119886119899=

8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

)

(2119899 + 3)

119887119899=

4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2

(2119899 + 3)

(28)

Then it is not difficult to verify that

119886119899=(16119899 + 20) + (16119899 + 12) 3

2119899+2

(2119899 + 3)gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

= minus119888119899

119889119899

lt 0

(29)

for all 119899 ge 0 where 119888119899= (256119899 + 48)3

2119899+2times 52119899+3

+ (384119899 +

464)52119899+3

+ (20481198992+7040119899+6000)3

2119899+2minus16 sdot 3

4119899+7+64 and

119889119899= [52119899+5

+(8119899+17)32119899+4

minus(8119899+22)][52119899+3

+(8119899+9)32119899+2

minus

(8119899 + 14)]Note that

1205933(0+) =

1198860

1198870

=2

3 120593

3 (infin) = 0 (30)

It follows from Lemma 2 and (27)ndash(29) that the func-tion 1198911015840

2(119909)1198921015840

2(119909) is strictly decreasing on (0infin) Therefore

Lemma 5 follows from (26) and (30) together with Lemma 1and the monotonicity of 1198911015840

2(119909)1198921015840

2(119909)

Lemma 6 The function

1205934 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909))]log [cos (119909)]

(31)

is strictly decreasing from (0 1205872) onto (0 13)

Proof Let1198913(119909) = log[(119909+sin(119909) cos(119909))(2 sin(119909))]119892

3(119909) =

log[cos(119909)] 1198914(119909) = cos2(119909)[119909 minus sin(119909) cos(119909)] and 119892

4(119909) =

sin2(119909)[119909 + sin(119909) cos(119909)] Then simple computations lead to

1205934 (119909) =

1198913 (119909)

1198923 (119909)

=1198913 (119909) minus 1198913 (0

+)

1198923 (119909) minus 1198923 (0)

(32)

1198911015840

3(119909)

1198921015840

3(119909)

=1198914 (119909)

1198924 (119909)

=1198914 (119909) minus 1198914 (0)

1198924 (119909) minus 1198924 (0)

(33)

1198911015840

4(119909)

1198921015840

4(119909)

= 1 minus1

12 + sin (2119909) (2119909) (34)

Abstract and Applied Analysis 5

Since the function 119909 rarr sin(119909)119909 is strictly decreasing on(0 120587) hence (34) leads to the conclusion that 1198911015840

4(119909)1198921015840

4(119909) is

strictly decreasing on (0 1205872)Note that

1205934(0+) = lim119909rarr0

+

1198911015840

4(119909)

1198921015840

4(119909)

=1

3 120593

4 (infin) = 0 (35)

Therefore Lemma 6 follows easily from (32) (33) (35)and Lemma 1 together with the monotonicity of 1198911015840

4(119909)1198921015840

4(119909)

Lemma 7 The function

1205935 (119909) =

log ((119909 + sinh (119909) cosh (119909)) (2 sinh (119909)))log [cosh (119909)]

(36)

is strictly increasing from (0 log(2 + radic3)) onto (13[2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1)

Proof Let1198915(119909) = log[(119909+sinh(119909) cosh(119909))(2 sinh(119909))] and

1198925(119909) = log[cosh(119909)] Then simple computations lead to

1205935 (119909) =

1198915 (119909)

1198925 (119909)

=1198915 (119909) minus 1198915 (0

+)

1198925 (119909) minus 1198925 (0)

(37)

1198911015840

5(119909)

1198921015840

5(119909)

=sinh (119909) cosh3 (119909) minus 119909cosh2 (119909)119909sinh2 (119909) + sinh3 (119909) cosh (119909)

=sinh (4119909) + 2 sinh (2119909) minus 4119909 cosh (2119909) minus 4119909sinh (4119909) minus 2 sinh (2119909) + 4119909 cosh (2119909) minus 4119909

=

suminfin

119899=1(22119899+2

(22119899minus 2119899) (2119899 + 1)) 119909

2119899+1

suminfin

119899=1(22119899+2 (22119899 + 2119899) (2119899 + 1)) 119909

2119899+1

=

suminfin

119899=0(22119899+4

[22119899+2

minus (2119899 + 2)] (2119899 + 3)) 1199092119899

suminfin

119899=0(22119899+4 [22119899+2 + (2119899 + 2)] (2119899 + 3)) 119909

2119899

(38)

Let

119886119899=

22119899+4

[22119899+2

minus (2119899 + 2)]

(2119899 + 3)

119887119899=

22119899+4

[22119899+2

+ (2119899 + 2)]

(2119899 + 3)

(39)

Then

119886119899gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

=(3119899 + 2) 2

2119899+2

(22119899+1 + 119899 + 1) (22119899+3 + 119899 + 2)gt 0

(40)

for all 119899 ge 0It follows from Lemma 2 and (38)ndash(40) that 1198911015840

5(119909)1198921015840

5(119909)

is strictly increasing on (0 log(2 + radic3))

Note that

1205935(0+) =

1198860

1198870

=1

3

1205935[log (2 + radic3)] =

2 log [2radic3 + log (2 + radic3)] minus log 32 log 2

minus 1

(41)

Therefore Lemma 7 follows from Lemma 1 (37) and (41)together with the monotonicity of 1198911015840

5(119909)1198921015840

5(119909)

Lemma 8 The function

1205936 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909) cos (119909))]log [sec (119909)]

(42)

is strictly increasing from (0 1205873) onto (23 [2 log(4120587+3radic3)minus2 log 6 minus log 3](2 log 2))

Proof Let 1198916(119909) = log[(119909 + sin(119909) cos(119909))(2 sin(119909) cos(119909))]

1198926(119909) = log[sec(119909)] 119891

7(119909) = sin(119909) cos(119909) minus 119909 cos(2119909) and

1198927(119909) = (119909+sin(119909) cos(119909))sin2(119909)Then simple computations

lead to

1205936 (119909) =

1198916 (119909)

1198926 (119909)

=1198916 (119909) minus 1198916 (0

+)

1198926 (119909) minus 1198926 (0)

1198911015840

6(119909)

1198921015840

6(119909)

=1198917 (119909)

1198927 (119909)

=1198917 (119909) minus 1198917 (0)

1198927 (119909) minus 1198927 (0)

1198911015840

7(119909)

1198921015840

7(119909)

=1

12 + sin (2119909) 2119909

(43)

Since the function 119909 rarr sin(2119909)(2119909) is strictly decreas-ing on (0 1205873) hence Lemma 1 and (43) lead to the conclu-sion that 120593

6(119909) is strictly increasing on (0 1205873)

Note that

1205936(0+) = lim119909rarr0

+

1198911015840

7(119909)

1198921015840

7(119909)

=2

3 (44)

1205936(120587

3) =

2 log (4120587 + 3radic3) minus 2 log 6 minus log 32 log 2

(45)

Therefore Lemma 8 follows easily from (44) and (45)together with the monotonicity of 120593

6(119909)

6 Abstract and Applied Analysis

3 Main Results

Theorem 9 The double inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

(46)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

(47)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

(48)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

(49)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge 1

1205722le 23 120573

2ge 1 120572

3le 13 120573

3ge [2 log(2radic3 + log(2 + radic3)) minus

log 3](2 log 2)minus1 = 04648 1205724le 23 and120573

4ge [2 log(4120587+

3radic3) minus 2 log 6 minus log 3](2 log 2) = 07733

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin) 119902 isin (0 1205872)119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be the parameters suchthat 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) = sec(119904) = 1 + V2Then (9)ndash(12) lead to the conclusion that inequalities (46)ndash(49) are respectively equivalent to

1205721lt 1 minus

log119873119860119867 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119901 + sinh (119901) cosh (119901)) (2 sinh (119901) cosh (119901))]

log (1 cosh (119901))

= 1 minus 1205933(119901) lt 120573

1

(50)

1205722lt 1 minus

log119873119867119860 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119902 + sin (119902) cos (119902)) (2 sin (119902))]

log [cos (119902)]

= 1 minus 1205934(119902) lt 120573

2

(51)

1205723ltlog119873119862119860 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log ((119903 + sinh (119903) cosh (119903)) (2 sinh (119903)))

log [cosh (119903)]

= 1205935 (119903) lt 1205733

(52)

1205724ltlog119873119860119862 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log [(119904 + sin (119904) cos (119904)) (2 sin (119904) cos (119904))]

log [sec (119904)]

= 1205936 (119904) lt 1205734

(53)

Therefore Theorem 9 follows easily from (50)ndash(53) andLemmas 5ndash8

Theorem 10 Let 1205725 1205735 1205726 1205736isin [0 12] Then the double

inequalities

119867[1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

(54)

hold for all 119886 119887 gt 0with 119886 = 119887 if and only if 1205725le 12minusradic66 =

009175 1205735ge 12 minusradic24 = 01464 120572

6le 12 minusradic36 =

02113 and 1205736ge 12 minus radic1 minus 12058742 = 02683

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120582 isin [0 12] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)and 119902 isin (0 1205872) be the parameters such that 1 cosh(119901) =cos(119902) = 1 minus V2 Then from (9) and (10) we have

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119860119867 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [1 +

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887)

2

times [1 minus2(1 minus 2120582)

2(cosh (119901) minus 1)

cosh (119901)minus

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)

times [1 +sinh (119901) minus 119901

sinh (119901) (cosh (119901) minus 1)minus 2(1 minus 2120582)

2]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)[1 + 120593

1(119901) minus 2(1 minus 2120582)

2]

(55)

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

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Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

be the Neuman means Then Neuman [11] proved that

119866 (119886 119887) lt 119871 (119886 119887) lt 119873119860119866 (119886 119887) lt 119875 (119886 119887) lt 119873119866119860 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119873119876119860 (119886 119887) lt 119879 (119886 119887)

lt 119873119860119876 (119886 119887) lt 119876 (119886 119887) lt 119862 (119886 119887)

(5)

for all 119886 119887 gt 0 with 119886 = 119887 and the double inequalities

1205721119860 (119886 119887) + (1 minus 1205721) 119866 (119886 119887)

lt 119873119866119860 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731) 119866 (119886 119887)

1205722119876 (119886 119887) + (1 minus 1205722) 119860 (119886 119887)

lt 119873119860119876 (119886 119887) lt 1205732119876 (119886 119887) + (1 minus 1205732) 119860 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723) 119866 (119886 119887)

lt 119873119860119866 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733) 119866 (119886 119887)

1205724119876 (119886 119887) + (1 minus 1205724) 119860 (119886 119887)

lt 119873119876119860 (119886 119887) lt 1205734119876 (119886 119887) + (1 minus 1205734) 119860 (119886 119887)

(6)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 23 120573

1ge

1205874 1205722le 23 120573

2ge (120587 minus 2)[4(radic2 minus 1)] = 0689 120572

3le 13

1205733ge 12 120572

4le 13 and 120573

4ge [log(1 +radic2) +radic2 minus 2][2(radic2 minus

1)] = 0356 Zhang et al [12] presented the best possible parameters

1205721 1205722 1205731 1205732isin [0 12] and 120572

3 1205724 1205733 1205734isin [12 1] such that

the double inequalities

119866 (1205721119886 + (1 minus 120572

1) 119887 1205721119887 + (1 minus 120572

1) 119886)

lt 119873119860119866 (119886 119887) lt 119866 (1205731119886 + (1 minus 1205731) 119887 1205731119887 + (1 minus 1205731) 119886)

119866 (1205722119886 + (1 minus 120572

2) 119887 1205722119887 + (1 minus 120572

2) 119886)

lt 119873119866119860 (119886 119887) lt 119866 (1205732119886 + (1 minus 1205732) 119887 1205732119887 + (1 minus 1205732) 119886)

119876 (1205723119886 + (1 minus 120572

3) 119887 1205723119887 + (1 minus 120572

3) 119886)

lt 119873119876119860 (119886 119887) lt 119876 (120573

3119886 + (1 minus 120573

3) 119887 1205733119887 + (1 minus 120573

3) 119886)

119876 (1205724119886 + (1 minus 120572

4) 119887 1205724119887 + (1 minus 120572

4) 119886)

lt 119873119860119876 (119886 119887) lt 119876 (120573

4119886 + (1 minus 120573

4) 119887 1205734119887 + (1 minus 120573

4) 119886)

(7)

hold for all 119886 119887 gt 0 with 119886 = 119887In [13] the authors found the greatest values 120572

1 1205722 1205723

1205724 1205725 1205726 1205727 and 120572

8and the least values 120573

1 1205732 1205733 1205734 1205735 1205736

1205737 and 120573

8such that the double inequalities

1198601205721(119886 119887) 119866

1minus1205721(119886 119887)

lt 119873119866119860 (119886 119887) lt 119860

1205731(119886 119887) 119866

1minus1205731(119886 119887)

1205722

119866 (119886 119887)+

1 minus 1205722

119860 (119886 119887)

lt1

119873119866119860 (119886 119887)

lt1205732

119866 (119886 119887)+

1 minus 1205732

119860 (119886 119887)

1198601205723(119886 119887) 119866

1minus1205723(119886 119887)

lt 119873119860119866 (119886 119887) lt 119860

1205733(119886 119887) 119866

1minus1205733(119886 119887)

1205724

119866 (119886 119887)+

1 minus 1205724

119860 (119886 119887)

lt1

119873119860119866 (119886 119887)

lt1205734

119866 (119886 119887)+

1 minus 1205734

119860 (119886 119887)

1198761205725(119886 119887) 119860

1minus1205725(119886 119887)

lt 119873119860119876 (119886 119887) lt 119876

1205735(119886 119887) 119860

1minus1205735(119886 119887)

1205726

119860 (119886 119887)+

1 minus 1205726

119876 (119886 119887)

lt1

119873119860119876 (119886 119887)

lt1205736

119860 (119886 119887)+

1 minus 1205736

119876 (119886 119887)

1198761205727(119886 119887) 119860

1minus1205727(119886 119887)

lt 119873119876119860 (119886 119887) lt 119876

1205737(119886 119887) 119860

1minus1205737(119886 119887)

1205728

119860 (119886 119887)+

1 minus 1205728

119876 (119886 119887)

lt1

119873119876119860 (119886 119887)

lt1205738

119860 (119886 119887)+

1 minus 1205738

119876 (119886 119887)

(8)

hold for all 119886 119887 gt 0 with 119886 = 119887Let 119886 gt 119887 gt 0 V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)

119902 isin (0 1205872) 119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be theparameters such that 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) =sec(119904) = 1 + V2 Then He et al [14] proved that

119873119860119867 (119886 119887) =

1

2119860 (119886 119887) [1 +

2119901

sinh (2119901)] (9)

119873119867119860 (119886 119887) =

1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)] (10)

119873119862119860 (119886 119887) =

1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)] (11)

119873119860119862 (119886 119887) =

1

2119860 (119886 119887) [1 +

2119904

sin (2119904)] (12)

119867(119886 119887) lt 119873119860119867 (119886 119887) lt 119873119867119860 (119886 119887) lt 119860 (119886 119887)

lt 119873119862119860 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(13)

Let 119909 isin [0 12] 119910 isin [12 1]119891(119909) = 119867[119909119886+(1minus119909)119887 119909119887+(1 minus 119909)119886] and 119892(119910) = 119862[119910119886 + (1 minus 119910)119887 119910119887 + (1 minus 119910)119886] Thenwe clearly see that

Abstract and Applied Analysis 3

119891 (0) = 119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119873119867119860 (119886 119887) lt 119860 (119886 119887)

= 119891(1

2) = 119892(

1

2) lt 119873

119862119860 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

= 119892 (1)

(14)

for all 119886 119887 gt 0 with 119886 = 119887 and the functions 119891 and 119892 arerespectively strictly increasing on the intervals [0 12] and[12 1] for fixed 119886 119887 gt 0 with 119886 = 119887

Themain purpose of this paper is to find the best possibleparameters 120572

1 1205722 1205723 1205724 1205731 1205732 1205733 1205734isin R 120572

5 1205726 1205735 1205736isin

[0 12] and 1205727 1205728 1205737 1205738isin [12 1] such that the double

inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

119867 [1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119867 [120573

7119886 + (1 minus 120573

7) 119887 1205737119887 + (1 minus 120573

7) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(15)

hold for all 119886 119887 gt 0 with 119886 = 119887 and present a new chain ofinequalities for certain bivariate means

2 Lemmas

In order to prove our main results we need several lemmaswhich we present in this section

Lemma 1 (see [15 Theorem 125]) For minusinfin lt 119886 lt 119887 lt infinlet 119891 119892 [119886 119887] rarr R be continuous on [119886 119887] and differentiable

on (119886 119887) let 1198921015840(119909) = 0 on (119886 119887) If 1198911015840(119909)1198921015840(119909) is increasing(decreasing) on (119886 119887) then so are

119891 (119909) minus 119891 (119886)

119892 (119909) minus 119892 (119886)

119891 (119909) minus 119891 (119887)

119892 (119909) minus 119892 (119887) (16)

If 1198911015840(119909)1198921015840(119909) is strictly monotone then the monotonicity inthe conclusion is also strict

Lemma 2 (see [16 Lemma 11]) Suppose that the power series119891(119909) = sum

infin

119899=0119886119899119909119899 and 119892(119909) = sum

infin

119899=0119887119899119909119899 have the radius of

convergence 119903 gt 0 and 119886119899 119887119899gt 0 for all 119899 ge 0 If the sequence

119886119899119887119899 is (strictly) increasing (decreasing) for all 119899 ge 0 then

the function 119891(119909)119892(119909) is also (strictly) increasing (decreasing)on (0 119903)

Lemma 3 The function

1205931 (119909) =

sinh (119909) minus 119909sinh (119909) [cosh (119909) minus 1]

(17)

is strictly decreasing from (0infin) onto (0 13)

Proof Making use of power series expansion we get

1205931 (119909) =

2 sinh (119909) minus 2119909sinh (2119909) minus 2 sinh (119909)

=suminfin

119899=1(2 (2119899 + 1)) 119909

2119899+1

suminfin

119899=1((22119899+1 minus 2) (2119899 + 1)) 119909

2119899+1

=suminfin

119899=0(2 (2119899 + 3)) 119909

2119899

suminfin

119899=0((22119899+3 minus 2) (2119899 + 3)) 119909

2119899

(18)

Let

119886119899=

2

(2119899 + 3) 119887

119899=22119899+3

minus 2

(2119899 + 3) (19)

Then

119886119899gt 0 119887

119899gt 0 (20)

and 119886119899119887119899= 1(2

2119899+2minus 1) is strictly decreasing for all 119899 ge 0

Note that

1205931(0+) =

1198860

1198870

=1

3 120593

1 (infin) = lim119899rarrinfin

119886119899

119887119899

= 0 (21)

Therefore Lemma 3 follows easily from Lemma 2 and(18)ndash(21) together with the monotonicity of the sequence119886119899119887119899

Lemma 4 The function

1205932 (119909) =

119909 minus sin (119909)sin (119909) [1 minus cos (119909)]

(22)

is strictly increasing from (0 1205872) onto (13 1205872 minus 1)

4 Abstract and Applied Analysis

Proof Let 1198911(119909) = 119909 minus sin(119909) and 119892

1(119909) = sin(119909)[1 minus cos(119909)]

Then simple computations lead to

1205932 (119909) =

1198911 (119909)

1198921 (119909)

=1198911 (119909) minus 1198911 (0)

1198921 (119909) minus 1198921 (0)

(23)

and 11989110158401(119909)1198921015840

1(119909) = 1[1 + 2 cos(119909)] is strictly increasing on

(0 1205872)Note that

1205932(0+) = lim119909rarr0

+

1198911015840

1(119909)

1198921015840

1(119909)

=1

3 120593

2(120587

2) =

120587

2minus 1 (24)

Therefore Lemma 4 follows from Lemma 1 (23) (24)and the monotonicity of 1198911015840

1(119909)1198921015840

1(119909)

Lemma 5 The function

1205933 (119909) =

log [(119909 + sinh (119909) cosh (119909)) (2 sinh (119909) cosh (119909))]log (1 cosh (119909))

(25)

is strictly decreasing from (0infin) onto (0 23)

Proof Let 1198912(119909) = log[(119909 + sinh(119909) cosh(119909))(2 sinh(119909)

cosh(119909))] and 1198922(119909) = log[1 cosh(119909)] Then simple compu-

tations lead to

1205933 (119909) =

1198912 (119909)

1198922 (119909)

=1198912 (119909) minus 1198912 (0

+)

1198922 (119909) minus 1198922 (0)

(26)

1198911015840

2(119909)

1198921015840

2(119909)

= (119909 [cosh3 (119909) + sinh2 (119909) cosh (119909)] minus sinh (119909) cosh2 (119909))

times (sinh2 (119909) cosh (119909) [119909 + sinh (119909) cosh (119909)])minus1

= (8119909 cosh (3119909) + 8119909 cosh (119909) minus 4 sinh (3119909) minus 4 sinh (119909))

times (4119909 cosh (3119909) minus 4119909 cosh (119909) + sinh (5119909) minus sinh (3119909)

minus2 sinh(119909))minus1

= (

infin

sum

119899=1

( (8 (2119899 + 1) (1 + 32119899) minus 4 (1 + 3

2119899+1))

times ((2119899 + 1))minus1) 1199092119899+1

)

times (

infin

sum

119899=1

( (4 (2119899 + 1) (32119899minus 1) + 5

2119899+1minus 32119899+1

minus 2)

times ((2119899 + 1))minus1) 1199092119899+1

)

minus1

= (

infin

sum

119899=0

( (8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

))

times ((2119899 + 3))minus1) 1199092119899)

times (

infin

sum

119899=0

( (4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2)

times ((2119899 + 3))minus1) 1199092119899)

minus1

(27)

Let

119886119899=

8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

)

(2119899 + 3)

119887119899=

4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2

(2119899 + 3)

(28)

Then it is not difficult to verify that

119886119899=(16119899 + 20) + (16119899 + 12) 3

2119899+2

(2119899 + 3)gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

= minus119888119899

119889119899

lt 0

(29)

for all 119899 ge 0 where 119888119899= (256119899 + 48)3

2119899+2times 52119899+3

+ (384119899 +

464)52119899+3

+ (20481198992+7040119899+6000)3

2119899+2minus16 sdot 3

4119899+7+64 and

119889119899= [52119899+5

+(8119899+17)32119899+4

minus(8119899+22)][52119899+3

+(8119899+9)32119899+2

minus

(8119899 + 14)]Note that

1205933(0+) =

1198860

1198870

=2

3 120593

3 (infin) = 0 (30)

It follows from Lemma 2 and (27)ndash(29) that the func-tion 1198911015840

2(119909)1198921015840

2(119909) is strictly decreasing on (0infin) Therefore

Lemma 5 follows from (26) and (30) together with Lemma 1and the monotonicity of 1198911015840

2(119909)1198921015840

2(119909)

Lemma 6 The function

1205934 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909))]log [cos (119909)]

(31)

is strictly decreasing from (0 1205872) onto (0 13)

Proof Let1198913(119909) = log[(119909+sin(119909) cos(119909))(2 sin(119909))]119892

3(119909) =

log[cos(119909)] 1198914(119909) = cos2(119909)[119909 minus sin(119909) cos(119909)] and 119892

4(119909) =

sin2(119909)[119909 + sin(119909) cos(119909)] Then simple computations lead to

1205934 (119909) =

1198913 (119909)

1198923 (119909)

=1198913 (119909) minus 1198913 (0

+)

1198923 (119909) minus 1198923 (0)

(32)

1198911015840

3(119909)

1198921015840

3(119909)

=1198914 (119909)

1198924 (119909)

=1198914 (119909) minus 1198914 (0)

1198924 (119909) minus 1198924 (0)

(33)

1198911015840

4(119909)

1198921015840

4(119909)

= 1 minus1

12 + sin (2119909) (2119909) (34)

Abstract and Applied Analysis 5

Since the function 119909 rarr sin(119909)119909 is strictly decreasing on(0 120587) hence (34) leads to the conclusion that 1198911015840

4(119909)1198921015840

4(119909) is

strictly decreasing on (0 1205872)Note that

1205934(0+) = lim119909rarr0

+

1198911015840

4(119909)

1198921015840

4(119909)

=1

3 120593

4 (infin) = 0 (35)

Therefore Lemma 6 follows easily from (32) (33) (35)and Lemma 1 together with the monotonicity of 1198911015840

4(119909)1198921015840

4(119909)

Lemma 7 The function

1205935 (119909) =

log ((119909 + sinh (119909) cosh (119909)) (2 sinh (119909)))log [cosh (119909)]

(36)

is strictly increasing from (0 log(2 + radic3)) onto (13[2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1)

Proof Let1198915(119909) = log[(119909+sinh(119909) cosh(119909))(2 sinh(119909))] and

1198925(119909) = log[cosh(119909)] Then simple computations lead to

1205935 (119909) =

1198915 (119909)

1198925 (119909)

=1198915 (119909) minus 1198915 (0

+)

1198925 (119909) minus 1198925 (0)

(37)

1198911015840

5(119909)

1198921015840

5(119909)

=sinh (119909) cosh3 (119909) minus 119909cosh2 (119909)119909sinh2 (119909) + sinh3 (119909) cosh (119909)

=sinh (4119909) + 2 sinh (2119909) minus 4119909 cosh (2119909) minus 4119909sinh (4119909) minus 2 sinh (2119909) + 4119909 cosh (2119909) minus 4119909

=

suminfin

119899=1(22119899+2

(22119899minus 2119899) (2119899 + 1)) 119909

2119899+1

suminfin

119899=1(22119899+2 (22119899 + 2119899) (2119899 + 1)) 119909

2119899+1

=

suminfin

119899=0(22119899+4

[22119899+2

minus (2119899 + 2)] (2119899 + 3)) 1199092119899

suminfin

119899=0(22119899+4 [22119899+2 + (2119899 + 2)] (2119899 + 3)) 119909

2119899

(38)

Let

119886119899=

22119899+4

[22119899+2

minus (2119899 + 2)]

(2119899 + 3)

119887119899=

22119899+4

[22119899+2

+ (2119899 + 2)]

(2119899 + 3)

(39)

Then

119886119899gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

=(3119899 + 2) 2

2119899+2

(22119899+1 + 119899 + 1) (22119899+3 + 119899 + 2)gt 0

(40)

for all 119899 ge 0It follows from Lemma 2 and (38)ndash(40) that 1198911015840

5(119909)1198921015840

5(119909)

is strictly increasing on (0 log(2 + radic3))

Note that

1205935(0+) =

1198860

1198870

=1

3

1205935[log (2 + radic3)] =

2 log [2radic3 + log (2 + radic3)] minus log 32 log 2

minus 1

(41)

Therefore Lemma 7 follows from Lemma 1 (37) and (41)together with the monotonicity of 1198911015840

5(119909)1198921015840

5(119909)

Lemma 8 The function

1205936 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909) cos (119909))]log [sec (119909)]

(42)

is strictly increasing from (0 1205873) onto (23 [2 log(4120587+3radic3)minus2 log 6 minus log 3](2 log 2))

Proof Let 1198916(119909) = log[(119909 + sin(119909) cos(119909))(2 sin(119909) cos(119909))]

1198926(119909) = log[sec(119909)] 119891

7(119909) = sin(119909) cos(119909) minus 119909 cos(2119909) and

1198927(119909) = (119909+sin(119909) cos(119909))sin2(119909)Then simple computations

lead to

1205936 (119909) =

1198916 (119909)

1198926 (119909)

=1198916 (119909) minus 1198916 (0

+)

1198926 (119909) minus 1198926 (0)

1198911015840

6(119909)

1198921015840

6(119909)

=1198917 (119909)

1198927 (119909)

=1198917 (119909) minus 1198917 (0)

1198927 (119909) minus 1198927 (0)

1198911015840

7(119909)

1198921015840

7(119909)

=1

12 + sin (2119909) 2119909

(43)

Since the function 119909 rarr sin(2119909)(2119909) is strictly decreas-ing on (0 1205873) hence Lemma 1 and (43) lead to the conclu-sion that 120593

6(119909) is strictly increasing on (0 1205873)

Note that

1205936(0+) = lim119909rarr0

+

1198911015840

7(119909)

1198921015840

7(119909)

=2

3 (44)

1205936(120587

3) =

2 log (4120587 + 3radic3) minus 2 log 6 minus log 32 log 2

(45)

Therefore Lemma 8 follows easily from (44) and (45)together with the monotonicity of 120593

6(119909)

6 Abstract and Applied Analysis

3 Main Results

Theorem 9 The double inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

(46)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

(47)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

(48)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

(49)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge 1

1205722le 23 120573

2ge 1 120572

3le 13 120573

3ge [2 log(2radic3 + log(2 + radic3)) minus

log 3](2 log 2)minus1 = 04648 1205724le 23 and120573

4ge [2 log(4120587+

3radic3) minus 2 log 6 minus log 3](2 log 2) = 07733

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin) 119902 isin (0 1205872)119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be the parameters suchthat 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) = sec(119904) = 1 + V2Then (9)ndash(12) lead to the conclusion that inequalities (46)ndash(49) are respectively equivalent to

1205721lt 1 minus

log119873119860119867 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119901 + sinh (119901) cosh (119901)) (2 sinh (119901) cosh (119901))]

log (1 cosh (119901))

= 1 minus 1205933(119901) lt 120573

1

(50)

1205722lt 1 minus

log119873119867119860 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119902 + sin (119902) cos (119902)) (2 sin (119902))]

log [cos (119902)]

= 1 minus 1205934(119902) lt 120573

2

(51)

1205723ltlog119873119862119860 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log ((119903 + sinh (119903) cosh (119903)) (2 sinh (119903)))

log [cosh (119903)]

= 1205935 (119903) lt 1205733

(52)

1205724ltlog119873119860119862 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log [(119904 + sin (119904) cos (119904)) (2 sin (119904) cos (119904))]

log [sec (119904)]

= 1205936 (119904) lt 1205734

(53)

Therefore Theorem 9 follows easily from (50)ndash(53) andLemmas 5ndash8

Theorem 10 Let 1205725 1205735 1205726 1205736isin [0 12] Then the double

inequalities

119867[1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

(54)

hold for all 119886 119887 gt 0with 119886 = 119887 if and only if 1205725le 12minusradic66 =

009175 1205735ge 12 minusradic24 = 01464 120572

6le 12 minusradic36 =

02113 and 1205736ge 12 minus radic1 minus 12058742 = 02683

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120582 isin [0 12] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)and 119902 isin (0 1205872) be the parameters such that 1 cosh(119901) =cos(119902) = 1 minus V2 Then from (9) and (10) we have

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119860119867 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [1 +

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887)

2

times [1 minus2(1 minus 2120582)

2(cosh (119901) minus 1)

cosh (119901)minus

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)

times [1 +sinh (119901) minus 119901

sinh (119901) (cosh (119901) minus 1)minus 2(1 minus 2120582)

2]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)[1 + 120593

1(119901) minus 2(1 minus 2120582)

2]

(55)

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

119891 (0) = 119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119873119867119860 (119886 119887) lt 119860 (119886 119887)

= 119891(1

2) = 119892(

1

2) lt 119873

119862119860 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

= 119892 (1)

(14)

for all 119886 119887 gt 0 with 119886 = 119887 and the functions 119891 and 119892 arerespectively strictly increasing on the intervals [0 12] and[12 1] for fixed 119886 119887 gt 0 with 119886 = 119887

Themain purpose of this paper is to find the best possibleparameters 120572

1 1205722 1205723 1205724 1205731 1205732 1205733 1205734isin R 120572

5 1205726 1205735 1205736isin

[0 12] and 1205727 1205728 1205737 1205738isin [12 1] such that the double

inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

119867 [1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119867 [120573

7119886 + (1 minus 120573

7) 119887 1205737119887 + (1 minus 120573

7) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(15)

hold for all 119886 119887 gt 0 with 119886 = 119887 and present a new chain ofinequalities for certain bivariate means

2 Lemmas

In order to prove our main results we need several lemmaswhich we present in this section

Lemma 1 (see [15 Theorem 125]) For minusinfin lt 119886 lt 119887 lt infinlet 119891 119892 [119886 119887] rarr R be continuous on [119886 119887] and differentiable

on (119886 119887) let 1198921015840(119909) = 0 on (119886 119887) If 1198911015840(119909)1198921015840(119909) is increasing(decreasing) on (119886 119887) then so are

119891 (119909) minus 119891 (119886)

119892 (119909) minus 119892 (119886)

119891 (119909) minus 119891 (119887)

119892 (119909) minus 119892 (119887) (16)

If 1198911015840(119909)1198921015840(119909) is strictly monotone then the monotonicity inthe conclusion is also strict

Lemma 2 (see [16 Lemma 11]) Suppose that the power series119891(119909) = sum

infin

119899=0119886119899119909119899 and 119892(119909) = sum

infin

119899=0119887119899119909119899 have the radius of

convergence 119903 gt 0 and 119886119899 119887119899gt 0 for all 119899 ge 0 If the sequence

119886119899119887119899 is (strictly) increasing (decreasing) for all 119899 ge 0 then

the function 119891(119909)119892(119909) is also (strictly) increasing (decreasing)on (0 119903)

Lemma 3 The function

1205931 (119909) =

sinh (119909) minus 119909sinh (119909) [cosh (119909) minus 1]

(17)

is strictly decreasing from (0infin) onto (0 13)

Proof Making use of power series expansion we get

1205931 (119909) =

2 sinh (119909) minus 2119909sinh (2119909) minus 2 sinh (119909)

=suminfin

119899=1(2 (2119899 + 1)) 119909

2119899+1

suminfin

119899=1((22119899+1 minus 2) (2119899 + 1)) 119909

2119899+1

=suminfin

119899=0(2 (2119899 + 3)) 119909

2119899

suminfin

119899=0((22119899+3 minus 2) (2119899 + 3)) 119909

2119899

(18)

Let

119886119899=

2

(2119899 + 3) 119887

119899=22119899+3

minus 2

(2119899 + 3) (19)

Then

119886119899gt 0 119887

119899gt 0 (20)

and 119886119899119887119899= 1(2

2119899+2minus 1) is strictly decreasing for all 119899 ge 0

Note that

1205931(0+) =

1198860

1198870

=1

3 120593

1 (infin) = lim119899rarrinfin

119886119899

119887119899

= 0 (21)

Therefore Lemma 3 follows easily from Lemma 2 and(18)ndash(21) together with the monotonicity of the sequence119886119899119887119899

Lemma 4 The function

1205932 (119909) =

119909 minus sin (119909)sin (119909) [1 minus cos (119909)]

(22)

is strictly increasing from (0 1205872) onto (13 1205872 minus 1)

4 Abstract and Applied Analysis

Proof Let 1198911(119909) = 119909 minus sin(119909) and 119892

1(119909) = sin(119909)[1 minus cos(119909)]

Then simple computations lead to

1205932 (119909) =

1198911 (119909)

1198921 (119909)

=1198911 (119909) minus 1198911 (0)

1198921 (119909) minus 1198921 (0)

(23)

and 11989110158401(119909)1198921015840

1(119909) = 1[1 + 2 cos(119909)] is strictly increasing on

(0 1205872)Note that

1205932(0+) = lim119909rarr0

+

1198911015840

1(119909)

1198921015840

1(119909)

=1

3 120593

2(120587

2) =

120587

2minus 1 (24)

Therefore Lemma 4 follows from Lemma 1 (23) (24)and the monotonicity of 1198911015840

1(119909)1198921015840

1(119909)

Lemma 5 The function

1205933 (119909) =

log [(119909 + sinh (119909) cosh (119909)) (2 sinh (119909) cosh (119909))]log (1 cosh (119909))

(25)

is strictly decreasing from (0infin) onto (0 23)

Proof Let 1198912(119909) = log[(119909 + sinh(119909) cosh(119909))(2 sinh(119909)

cosh(119909))] and 1198922(119909) = log[1 cosh(119909)] Then simple compu-

tations lead to

1205933 (119909) =

1198912 (119909)

1198922 (119909)

=1198912 (119909) minus 1198912 (0

+)

1198922 (119909) minus 1198922 (0)

(26)

1198911015840

2(119909)

1198921015840

2(119909)

= (119909 [cosh3 (119909) + sinh2 (119909) cosh (119909)] minus sinh (119909) cosh2 (119909))

times (sinh2 (119909) cosh (119909) [119909 + sinh (119909) cosh (119909)])minus1

= (8119909 cosh (3119909) + 8119909 cosh (119909) minus 4 sinh (3119909) minus 4 sinh (119909))

times (4119909 cosh (3119909) minus 4119909 cosh (119909) + sinh (5119909) minus sinh (3119909)

minus2 sinh(119909))minus1

= (

infin

sum

119899=1

( (8 (2119899 + 1) (1 + 32119899) minus 4 (1 + 3

2119899+1))

times ((2119899 + 1))minus1) 1199092119899+1

)

times (

infin

sum

119899=1

( (4 (2119899 + 1) (32119899minus 1) + 5

2119899+1minus 32119899+1

minus 2)

times ((2119899 + 1))minus1) 1199092119899+1

)

minus1

= (

infin

sum

119899=0

( (8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

))

times ((2119899 + 3))minus1) 1199092119899)

times (

infin

sum

119899=0

( (4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2)

times ((2119899 + 3))minus1) 1199092119899)

minus1

(27)

Let

119886119899=

8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

)

(2119899 + 3)

119887119899=

4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2

(2119899 + 3)

(28)

Then it is not difficult to verify that

119886119899=(16119899 + 20) + (16119899 + 12) 3

2119899+2

(2119899 + 3)gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

= minus119888119899

119889119899

lt 0

(29)

for all 119899 ge 0 where 119888119899= (256119899 + 48)3

2119899+2times 52119899+3

+ (384119899 +

464)52119899+3

+ (20481198992+7040119899+6000)3

2119899+2minus16 sdot 3

4119899+7+64 and

119889119899= [52119899+5

+(8119899+17)32119899+4

minus(8119899+22)][52119899+3

+(8119899+9)32119899+2

minus

(8119899 + 14)]Note that

1205933(0+) =

1198860

1198870

=2

3 120593

3 (infin) = 0 (30)

It follows from Lemma 2 and (27)ndash(29) that the func-tion 1198911015840

2(119909)1198921015840

2(119909) is strictly decreasing on (0infin) Therefore

Lemma 5 follows from (26) and (30) together with Lemma 1and the monotonicity of 1198911015840

2(119909)1198921015840

2(119909)

Lemma 6 The function

1205934 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909))]log [cos (119909)]

(31)

is strictly decreasing from (0 1205872) onto (0 13)

Proof Let1198913(119909) = log[(119909+sin(119909) cos(119909))(2 sin(119909))]119892

3(119909) =

log[cos(119909)] 1198914(119909) = cos2(119909)[119909 minus sin(119909) cos(119909)] and 119892

4(119909) =

sin2(119909)[119909 + sin(119909) cos(119909)] Then simple computations lead to

1205934 (119909) =

1198913 (119909)

1198923 (119909)

=1198913 (119909) minus 1198913 (0

+)

1198923 (119909) minus 1198923 (0)

(32)

1198911015840

3(119909)

1198921015840

3(119909)

=1198914 (119909)

1198924 (119909)

=1198914 (119909) minus 1198914 (0)

1198924 (119909) minus 1198924 (0)

(33)

1198911015840

4(119909)

1198921015840

4(119909)

= 1 minus1

12 + sin (2119909) (2119909) (34)

Abstract and Applied Analysis 5

Since the function 119909 rarr sin(119909)119909 is strictly decreasing on(0 120587) hence (34) leads to the conclusion that 1198911015840

4(119909)1198921015840

4(119909) is

strictly decreasing on (0 1205872)Note that

1205934(0+) = lim119909rarr0

+

1198911015840

4(119909)

1198921015840

4(119909)

=1

3 120593

4 (infin) = 0 (35)

Therefore Lemma 6 follows easily from (32) (33) (35)and Lemma 1 together with the monotonicity of 1198911015840

4(119909)1198921015840

4(119909)

Lemma 7 The function

1205935 (119909) =

log ((119909 + sinh (119909) cosh (119909)) (2 sinh (119909)))log [cosh (119909)]

(36)

is strictly increasing from (0 log(2 + radic3)) onto (13[2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1)

Proof Let1198915(119909) = log[(119909+sinh(119909) cosh(119909))(2 sinh(119909))] and

1198925(119909) = log[cosh(119909)] Then simple computations lead to

1205935 (119909) =

1198915 (119909)

1198925 (119909)

=1198915 (119909) minus 1198915 (0

+)

1198925 (119909) minus 1198925 (0)

(37)

1198911015840

5(119909)

1198921015840

5(119909)

=sinh (119909) cosh3 (119909) minus 119909cosh2 (119909)119909sinh2 (119909) + sinh3 (119909) cosh (119909)

=sinh (4119909) + 2 sinh (2119909) minus 4119909 cosh (2119909) minus 4119909sinh (4119909) minus 2 sinh (2119909) + 4119909 cosh (2119909) minus 4119909

=

suminfin

119899=1(22119899+2

(22119899minus 2119899) (2119899 + 1)) 119909

2119899+1

suminfin

119899=1(22119899+2 (22119899 + 2119899) (2119899 + 1)) 119909

2119899+1

=

suminfin

119899=0(22119899+4

[22119899+2

minus (2119899 + 2)] (2119899 + 3)) 1199092119899

suminfin

119899=0(22119899+4 [22119899+2 + (2119899 + 2)] (2119899 + 3)) 119909

2119899

(38)

Let

119886119899=

22119899+4

[22119899+2

minus (2119899 + 2)]

(2119899 + 3)

119887119899=

22119899+4

[22119899+2

+ (2119899 + 2)]

(2119899 + 3)

(39)

Then

119886119899gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

=(3119899 + 2) 2

2119899+2

(22119899+1 + 119899 + 1) (22119899+3 + 119899 + 2)gt 0

(40)

for all 119899 ge 0It follows from Lemma 2 and (38)ndash(40) that 1198911015840

5(119909)1198921015840

5(119909)

is strictly increasing on (0 log(2 + radic3))

Note that

1205935(0+) =

1198860

1198870

=1

3

1205935[log (2 + radic3)] =

2 log [2radic3 + log (2 + radic3)] minus log 32 log 2

minus 1

(41)

Therefore Lemma 7 follows from Lemma 1 (37) and (41)together with the monotonicity of 1198911015840

5(119909)1198921015840

5(119909)

Lemma 8 The function

1205936 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909) cos (119909))]log [sec (119909)]

(42)

is strictly increasing from (0 1205873) onto (23 [2 log(4120587+3radic3)minus2 log 6 minus log 3](2 log 2))

Proof Let 1198916(119909) = log[(119909 + sin(119909) cos(119909))(2 sin(119909) cos(119909))]

1198926(119909) = log[sec(119909)] 119891

7(119909) = sin(119909) cos(119909) minus 119909 cos(2119909) and

1198927(119909) = (119909+sin(119909) cos(119909))sin2(119909)Then simple computations

lead to

1205936 (119909) =

1198916 (119909)

1198926 (119909)

=1198916 (119909) minus 1198916 (0

+)

1198926 (119909) minus 1198926 (0)

1198911015840

6(119909)

1198921015840

6(119909)

=1198917 (119909)

1198927 (119909)

=1198917 (119909) minus 1198917 (0)

1198927 (119909) minus 1198927 (0)

1198911015840

7(119909)

1198921015840

7(119909)

=1

12 + sin (2119909) 2119909

(43)

Since the function 119909 rarr sin(2119909)(2119909) is strictly decreas-ing on (0 1205873) hence Lemma 1 and (43) lead to the conclu-sion that 120593

6(119909) is strictly increasing on (0 1205873)

Note that

1205936(0+) = lim119909rarr0

+

1198911015840

7(119909)

1198921015840

7(119909)

=2

3 (44)

1205936(120587

3) =

2 log (4120587 + 3radic3) minus 2 log 6 minus log 32 log 2

(45)

Therefore Lemma 8 follows easily from (44) and (45)together with the monotonicity of 120593

6(119909)

6 Abstract and Applied Analysis

3 Main Results

Theorem 9 The double inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

(46)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

(47)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

(48)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

(49)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge 1

1205722le 23 120573

2ge 1 120572

3le 13 120573

3ge [2 log(2radic3 + log(2 + radic3)) minus

log 3](2 log 2)minus1 = 04648 1205724le 23 and120573

4ge [2 log(4120587+

3radic3) minus 2 log 6 minus log 3](2 log 2) = 07733

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin) 119902 isin (0 1205872)119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be the parameters suchthat 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) = sec(119904) = 1 + V2Then (9)ndash(12) lead to the conclusion that inequalities (46)ndash(49) are respectively equivalent to

1205721lt 1 minus

log119873119860119867 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119901 + sinh (119901) cosh (119901)) (2 sinh (119901) cosh (119901))]

log (1 cosh (119901))

= 1 minus 1205933(119901) lt 120573

1

(50)

1205722lt 1 minus

log119873119867119860 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119902 + sin (119902) cos (119902)) (2 sin (119902))]

log [cos (119902)]

= 1 minus 1205934(119902) lt 120573

2

(51)

1205723ltlog119873119862119860 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log ((119903 + sinh (119903) cosh (119903)) (2 sinh (119903)))

log [cosh (119903)]

= 1205935 (119903) lt 1205733

(52)

1205724ltlog119873119860119862 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log [(119904 + sin (119904) cos (119904)) (2 sin (119904) cos (119904))]

log [sec (119904)]

= 1205936 (119904) lt 1205734

(53)

Therefore Theorem 9 follows easily from (50)ndash(53) andLemmas 5ndash8

Theorem 10 Let 1205725 1205735 1205726 1205736isin [0 12] Then the double

inequalities

119867[1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

(54)

hold for all 119886 119887 gt 0with 119886 = 119887 if and only if 1205725le 12minusradic66 =

009175 1205735ge 12 minusradic24 = 01464 120572

6le 12 minusradic36 =

02113 and 1205736ge 12 minus radic1 minus 12058742 = 02683

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120582 isin [0 12] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)and 119902 isin (0 1205872) be the parameters such that 1 cosh(119901) =cos(119902) = 1 minus V2 Then from (9) and (10) we have

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119860119867 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [1 +

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887)

2

times [1 minus2(1 minus 2120582)

2(cosh (119901) minus 1)

cosh (119901)minus

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)

times [1 +sinh (119901) minus 119901

sinh (119901) (cosh (119901) minus 1)minus 2(1 minus 2120582)

2]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)[1 + 120593

1(119901) minus 2(1 minus 2120582)

2]

(55)

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Proof Let 1198911(119909) = 119909 minus sin(119909) and 119892

1(119909) = sin(119909)[1 minus cos(119909)]

Then simple computations lead to

1205932 (119909) =

1198911 (119909)

1198921 (119909)

=1198911 (119909) minus 1198911 (0)

1198921 (119909) minus 1198921 (0)

(23)

and 11989110158401(119909)1198921015840

1(119909) = 1[1 + 2 cos(119909)] is strictly increasing on

(0 1205872)Note that

1205932(0+) = lim119909rarr0

+

1198911015840

1(119909)

1198921015840

1(119909)

=1

3 120593

2(120587

2) =

120587

2minus 1 (24)

Therefore Lemma 4 follows from Lemma 1 (23) (24)and the monotonicity of 1198911015840

1(119909)1198921015840

1(119909)

Lemma 5 The function

1205933 (119909) =

log [(119909 + sinh (119909) cosh (119909)) (2 sinh (119909) cosh (119909))]log (1 cosh (119909))

(25)

is strictly decreasing from (0infin) onto (0 23)

Proof Let 1198912(119909) = log[(119909 + sinh(119909) cosh(119909))(2 sinh(119909)

cosh(119909))] and 1198922(119909) = log[1 cosh(119909)] Then simple compu-

tations lead to

1205933 (119909) =

1198912 (119909)

1198922 (119909)

=1198912 (119909) minus 1198912 (0

+)

1198922 (119909) minus 1198922 (0)

(26)

1198911015840

2(119909)

1198921015840

2(119909)

= (119909 [cosh3 (119909) + sinh2 (119909) cosh (119909)] minus sinh (119909) cosh2 (119909))

times (sinh2 (119909) cosh (119909) [119909 + sinh (119909) cosh (119909)])minus1

= (8119909 cosh (3119909) + 8119909 cosh (119909) minus 4 sinh (3119909) minus 4 sinh (119909))

times (4119909 cosh (3119909) minus 4119909 cosh (119909) + sinh (5119909) minus sinh (3119909)

minus2 sinh(119909))minus1

= (

infin

sum

119899=1

( (8 (2119899 + 1) (1 + 32119899) minus 4 (1 + 3

2119899+1))

times ((2119899 + 1))minus1) 1199092119899+1

)

times (

infin

sum

119899=1

( (4 (2119899 + 1) (32119899minus 1) + 5

2119899+1minus 32119899+1

minus 2)

times ((2119899 + 1))minus1) 1199092119899+1

)

minus1

= (

infin

sum

119899=0

( (8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

))

times ((2119899 + 3))minus1) 1199092119899)

times (

infin

sum

119899=0

( (4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2)

times ((2119899 + 3))minus1) 1199092119899)

minus1

(27)

Let

119886119899=

8 (2119899 + 3) (1 + 32119899+2

) minus 4 (1 + 32119899+3

)

(2119899 + 3)

119887119899=

4 (2119899 + 3) (32119899+2

minus 1) + 52119899+3

minus 32119899+3

minus 2

(2119899 + 3)

(28)

Then it is not difficult to verify that

119886119899=(16119899 + 20) + (16119899 + 12) 3

2119899+2

(2119899 + 3)gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

= minus119888119899

119889119899

lt 0

(29)

for all 119899 ge 0 where 119888119899= (256119899 + 48)3

2119899+2times 52119899+3

+ (384119899 +

464)52119899+3

+ (20481198992+7040119899+6000)3

2119899+2minus16 sdot 3

4119899+7+64 and

119889119899= [52119899+5

+(8119899+17)32119899+4

minus(8119899+22)][52119899+3

+(8119899+9)32119899+2

minus

(8119899 + 14)]Note that

1205933(0+) =

1198860

1198870

=2

3 120593

3 (infin) = 0 (30)

It follows from Lemma 2 and (27)ndash(29) that the func-tion 1198911015840

2(119909)1198921015840

2(119909) is strictly decreasing on (0infin) Therefore

Lemma 5 follows from (26) and (30) together with Lemma 1and the monotonicity of 1198911015840

2(119909)1198921015840

2(119909)

Lemma 6 The function

1205934 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909))]log [cos (119909)]

(31)

is strictly decreasing from (0 1205872) onto (0 13)

Proof Let1198913(119909) = log[(119909+sin(119909) cos(119909))(2 sin(119909))]119892

3(119909) =

log[cos(119909)] 1198914(119909) = cos2(119909)[119909 minus sin(119909) cos(119909)] and 119892

4(119909) =

sin2(119909)[119909 + sin(119909) cos(119909)] Then simple computations lead to

1205934 (119909) =

1198913 (119909)

1198923 (119909)

=1198913 (119909) minus 1198913 (0

+)

1198923 (119909) minus 1198923 (0)

(32)

1198911015840

3(119909)

1198921015840

3(119909)

=1198914 (119909)

1198924 (119909)

=1198914 (119909) minus 1198914 (0)

1198924 (119909) minus 1198924 (0)

(33)

1198911015840

4(119909)

1198921015840

4(119909)

= 1 minus1

12 + sin (2119909) (2119909) (34)

Abstract and Applied Analysis 5

Since the function 119909 rarr sin(119909)119909 is strictly decreasing on(0 120587) hence (34) leads to the conclusion that 1198911015840

4(119909)1198921015840

4(119909) is

strictly decreasing on (0 1205872)Note that

1205934(0+) = lim119909rarr0

+

1198911015840

4(119909)

1198921015840

4(119909)

=1

3 120593

4 (infin) = 0 (35)

Therefore Lemma 6 follows easily from (32) (33) (35)and Lemma 1 together with the monotonicity of 1198911015840

4(119909)1198921015840

4(119909)

Lemma 7 The function

1205935 (119909) =

log ((119909 + sinh (119909) cosh (119909)) (2 sinh (119909)))log [cosh (119909)]

(36)

is strictly increasing from (0 log(2 + radic3)) onto (13[2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1)

Proof Let1198915(119909) = log[(119909+sinh(119909) cosh(119909))(2 sinh(119909))] and

1198925(119909) = log[cosh(119909)] Then simple computations lead to

1205935 (119909) =

1198915 (119909)

1198925 (119909)

=1198915 (119909) minus 1198915 (0

+)

1198925 (119909) minus 1198925 (0)

(37)

1198911015840

5(119909)

1198921015840

5(119909)

=sinh (119909) cosh3 (119909) minus 119909cosh2 (119909)119909sinh2 (119909) + sinh3 (119909) cosh (119909)

=sinh (4119909) + 2 sinh (2119909) minus 4119909 cosh (2119909) minus 4119909sinh (4119909) minus 2 sinh (2119909) + 4119909 cosh (2119909) minus 4119909

=

suminfin

119899=1(22119899+2

(22119899minus 2119899) (2119899 + 1)) 119909

2119899+1

suminfin

119899=1(22119899+2 (22119899 + 2119899) (2119899 + 1)) 119909

2119899+1

=

suminfin

119899=0(22119899+4

[22119899+2

minus (2119899 + 2)] (2119899 + 3)) 1199092119899

suminfin

119899=0(22119899+4 [22119899+2 + (2119899 + 2)] (2119899 + 3)) 119909

2119899

(38)

Let

119886119899=

22119899+4

[22119899+2

minus (2119899 + 2)]

(2119899 + 3)

119887119899=

22119899+4

[22119899+2

+ (2119899 + 2)]

(2119899 + 3)

(39)

Then

119886119899gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

=(3119899 + 2) 2

2119899+2

(22119899+1 + 119899 + 1) (22119899+3 + 119899 + 2)gt 0

(40)

for all 119899 ge 0It follows from Lemma 2 and (38)ndash(40) that 1198911015840

5(119909)1198921015840

5(119909)

is strictly increasing on (0 log(2 + radic3))

Note that

1205935(0+) =

1198860

1198870

=1

3

1205935[log (2 + radic3)] =

2 log [2radic3 + log (2 + radic3)] minus log 32 log 2

minus 1

(41)

Therefore Lemma 7 follows from Lemma 1 (37) and (41)together with the monotonicity of 1198911015840

5(119909)1198921015840

5(119909)

Lemma 8 The function

1205936 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909) cos (119909))]log [sec (119909)]

(42)

is strictly increasing from (0 1205873) onto (23 [2 log(4120587+3radic3)minus2 log 6 minus log 3](2 log 2))

Proof Let 1198916(119909) = log[(119909 + sin(119909) cos(119909))(2 sin(119909) cos(119909))]

1198926(119909) = log[sec(119909)] 119891

7(119909) = sin(119909) cos(119909) minus 119909 cos(2119909) and

1198927(119909) = (119909+sin(119909) cos(119909))sin2(119909)Then simple computations

lead to

1205936 (119909) =

1198916 (119909)

1198926 (119909)

=1198916 (119909) minus 1198916 (0

+)

1198926 (119909) minus 1198926 (0)

1198911015840

6(119909)

1198921015840

6(119909)

=1198917 (119909)

1198927 (119909)

=1198917 (119909) minus 1198917 (0)

1198927 (119909) minus 1198927 (0)

1198911015840

7(119909)

1198921015840

7(119909)

=1

12 + sin (2119909) 2119909

(43)

Since the function 119909 rarr sin(2119909)(2119909) is strictly decreas-ing on (0 1205873) hence Lemma 1 and (43) lead to the conclu-sion that 120593

6(119909) is strictly increasing on (0 1205873)

Note that

1205936(0+) = lim119909rarr0

+

1198911015840

7(119909)

1198921015840

7(119909)

=2

3 (44)

1205936(120587

3) =

2 log (4120587 + 3radic3) minus 2 log 6 minus log 32 log 2

(45)

Therefore Lemma 8 follows easily from (44) and (45)together with the monotonicity of 120593

6(119909)

6 Abstract and Applied Analysis

3 Main Results

Theorem 9 The double inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

(46)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

(47)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

(48)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

(49)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge 1

1205722le 23 120573

2ge 1 120572

3le 13 120573

3ge [2 log(2radic3 + log(2 + radic3)) minus

log 3](2 log 2)minus1 = 04648 1205724le 23 and120573

4ge [2 log(4120587+

3radic3) minus 2 log 6 minus log 3](2 log 2) = 07733

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin) 119902 isin (0 1205872)119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be the parameters suchthat 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) = sec(119904) = 1 + V2Then (9)ndash(12) lead to the conclusion that inequalities (46)ndash(49) are respectively equivalent to

1205721lt 1 minus

log119873119860119867 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119901 + sinh (119901) cosh (119901)) (2 sinh (119901) cosh (119901))]

log (1 cosh (119901))

= 1 minus 1205933(119901) lt 120573

1

(50)

1205722lt 1 minus

log119873119867119860 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119902 + sin (119902) cos (119902)) (2 sin (119902))]

log [cos (119902)]

= 1 minus 1205934(119902) lt 120573

2

(51)

1205723ltlog119873119862119860 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log ((119903 + sinh (119903) cosh (119903)) (2 sinh (119903)))

log [cosh (119903)]

= 1205935 (119903) lt 1205733

(52)

1205724ltlog119873119860119862 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log [(119904 + sin (119904) cos (119904)) (2 sin (119904) cos (119904))]

log [sec (119904)]

= 1205936 (119904) lt 1205734

(53)

Therefore Theorem 9 follows easily from (50)ndash(53) andLemmas 5ndash8

Theorem 10 Let 1205725 1205735 1205726 1205736isin [0 12] Then the double

inequalities

119867[1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

(54)

hold for all 119886 119887 gt 0with 119886 = 119887 if and only if 1205725le 12minusradic66 =

009175 1205735ge 12 minusradic24 = 01464 120572

6le 12 minusradic36 =

02113 and 1205736ge 12 minus radic1 minus 12058742 = 02683

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120582 isin [0 12] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)and 119902 isin (0 1205872) be the parameters such that 1 cosh(119901) =cos(119902) = 1 minus V2 Then from (9) and (10) we have

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119860119867 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [1 +

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887)

2

times [1 minus2(1 minus 2120582)

2(cosh (119901) minus 1)

cosh (119901)minus

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)

times [1 +sinh (119901) minus 119901

sinh (119901) (cosh (119901) minus 1)minus 2(1 minus 2120582)

2]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)[1 + 120593

1(119901) minus 2(1 minus 2120582)

2]

(55)

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Since the function 119909 rarr sin(119909)119909 is strictly decreasing on(0 120587) hence (34) leads to the conclusion that 1198911015840

4(119909)1198921015840

4(119909) is

strictly decreasing on (0 1205872)Note that

1205934(0+) = lim119909rarr0

+

1198911015840

4(119909)

1198921015840

4(119909)

=1

3 120593

4 (infin) = 0 (35)

Therefore Lemma 6 follows easily from (32) (33) (35)and Lemma 1 together with the monotonicity of 1198911015840

4(119909)1198921015840

4(119909)

Lemma 7 The function

1205935 (119909) =

log ((119909 + sinh (119909) cosh (119909)) (2 sinh (119909)))log [cosh (119909)]

(36)

is strictly increasing from (0 log(2 + radic3)) onto (13[2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1)

Proof Let1198915(119909) = log[(119909+sinh(119909) cosh(119909))(2 sinh(119909))] and

1198925(119909) = log[cosh(119909)] Then simple computations lead to

1205935 (119909) =

1198915 (119909)

1198925 (119909)

=1198915 (119909) minus 1198915 (0

+)

1198925 (119909) minus 1198925 (0)

(37)

1198911015840

5(119909)

1198921015840

5(119909)

=sinh (119909) cosh3 (119909) minus 119909cosh2 (119909)119909sinh2 (119909) + sinh3 (119909) cosh (119909)

=sinh (4119909) + 2 sinh (2119909) minus 4119909 cosh (2119909) minus 4119909sinh (4119909) minus 2 sinh (2119909) + 4119909 cosh (2119909) minus 4119909

=

suminfin

119899=1(22119899+2

(22119899minus 2119899) (2119899 + 1)) 119909

2119899+1

suminfin

119899=1(22119899+2 (22119899 + 2119899) (2119899 + 1)) 119909

2119899+1

=

suminfin

119899=0(22119899+4

[22119899+2

minus (2119899 + 2)] (2119899 + 3)) 1199092119899

suminfin

119899=0(22119899+4 [22119899+2 + (2119899 + 2)] (2119899 + 3)) 119909

2119899

(38)

Let

119886119899=

22119899+4

[22119899+2

minus (2119899 + 2)]

(2119899 + 3)

119887119899=

22119899+4

[22119899+2

+ (2119899 + 2)]

(2119899 + 3)

(39)

Then

119886119899gt 0 119887

119899gt 0

119886119899+1

119887119899+1

minus119886119899

119887119899

=(3119899 + 2) 2

2119899+2

(22119899+1 + 119899 + 1) (22119899+3 + 119899 + 2)gt 0

(40)

for all 119899 ge 0It follows from Lemma 2 and (38)ndash(40) that 1198911015840

5(119909)1198921015840

5(119909)

is strictly increasing on (0 log(2 + radic3))

Note that

1205935(0+) =

1198860

1198870

=1

3

1205935[log (2 + radic3)] =

2 log [2radic3 + log (2 + radic3)] minus log 32 log 2

minus 1

(41)

Therefore Lemma 7 follows from Lemma 1 (37) and (41)together with the monotonicity of 1198911015840

5(119909)1198921015840

5(119909)

Lemma 8 The function

1205936 (119909) =

log [(119909 + sin (119909) cos (119909)) (2 sin (119909) cos (119909))]log [sec (119909)]

(42)

is strictly increasing from (0 1205873) onto (23 [2 log(4120587+3radic3)minus2 log 6 minus log 3](2 log 2))

Proof Let 1198916(119909) = log[(119909 + sin(119909) cos(119909))(2 sin(119909) cos(119909))]

1198926(119909) = log[sec(119909)] 119891

7(119909) = sin(119909) cos(119909) minus 119909 cos(2119909) and

1198927(119909) = (119909+sin(119909) cos(119909))sin2(119909)Then simple computations

lead to

1205936 (119909) =

1198916 (119909)

1198926 (119909)

=1198916 (119909) minus 1198916 (0

+)

1198926 (119909) minus 1198926 (0)

1198911015840

6(119909)

1198921015840

6(119909)

=1198917 (119909)

1198927 (119909)

=1198917 (119909) minus 1198917 (0)

1198927 (119909) minus 1198927 (0)

1198911015840

7(119909)

1198921015840

7(119909)

=1

12 + sin (2119909) 2119909

(43)

Since the function 119909 rarr sin(2119909)(2119909) is strictly decreas-ing on (0 1205873) hence Lemma 1 and (43) lead to the conclu-sion that 120593

6(119909) is strictly increasing on (0 1205873)

Note that

1205936(0+) = lim119909rarr0

+

1198911015840

7(119909)

1198921015840

7(119909)

=2

3 (44)

1205936(120587

3) =

2 log (4120587 + 3radic3) minus 2 log 6 minus log 32 log 2

(45)

Therefore Lemma 8 follows easily from (44) and (45)together with the monotonicity of 120593

6(119909)

6 Abstract and Applied Analysis

3 Main Results

Theorem 9 The double inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

(46)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

(47)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

(48)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

(49)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge 1

1205722le 23 120573

2ge 1 120572

3le 13 120573

3ge [2 log(2radic3 + log(2 + radic3)) minus

log 3](2 log 2)minus1 = 04648 1205724le 23 and120573

4ge [2 log(4120587+

3radic3) minus 2 log 6 minus log 3](2 log 2) = 07733

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin) 119902 isin (0 1205872)119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be the parameters suchthat 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) = sec(119904) = 1 + V2Then (9)ndash(12) lead to the conclusion that inequalities (46)ndash(49) are respectively equivalent to

1205721lt 1 minus

log119873119860119867 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119901 + sinh (119901) cosh (119901)) (2 sinh (119901) cosh (119901))]

log (1 cosh (119901))

= 1 minus 1205933(119901) lt 120573

1

(50)

1205722lt 1 minus

log119873119867119860 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119902 + sin (119902) cos (119902)) (2 sin (119902))]

log [cos (119902)]

= 1 minus 1205934(119902) lt 120573

2

(51)

1205723ltlog119873119862119860 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log ((119903 + sinh (119903) cosh (119903)) (2 sinh (119903)))

log [cosh (119903)]

= 1205935 (119903) lt 1205733

(52)

1205724ltlog119873119860119862 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log [(119904 + sin (119904) cos (119904)) (2 sin (119904) cos (119904))]

log [sec (119904)]

= 1205936 (119904) lt 1205734

(53)

Therefore Theorem 9 follows easily from (50)ndash(53) andLemmas 5ndash8

Theorem 10 Let 1205725 1205735 1205726 1205736isin [0 12] Then the double

inequalities

119867[1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

(54)

hold for all 119886 119887 gt 0with 119886 = 119887 if and only if 1205725le 12minusradic66 =

009175 1205735ge 12 minusradic24 = 01464 120572

6le 12 minusradic36 =

02113 and 1205736ge 12 minus radic1 minus 12058742 = 02683

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120582 isin [0 12] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)and 119902 isin (0 1205872) be the parameters such that 1 cosh(119901) =cos(119902) = 1 minus V2 Then from (9) and (10) we have

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119860119867 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [1 +

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887)

2

times [1 minus2(1 minus 2120582)

2(cosh (119901) minus 1)

cosh (119901)minus

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)

times [1 +sinh (119901) minus 119901

sinh (119901) (cosh (119901) minus 1)minus 2(1 minus 2120582)

2]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)[1 + 120593

1(119901) minus 2(1 minus 2120582)

2]

(55)

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

3 Main Results

Theorem 9 The double inequalities

1198601205721(119886 119887)119867

1minus1205721(119886 119887)

lt 119873119860119867 (119886 119887) lt 119860

1205731(119886 119887)119867

1minus1205731(119886 119887)

(46)

1198601205722(119886 119887)119867

1minus1205722(119886 119887)

lt 119873119867119860 (119886 119887) lt 119860

1205732(119886 119887)119867

1minus1205732(119886 119887)

(47)

1198621205723(119886 119887) 119860

1minus1205723(119886 119887)

lt 119873119862119860 (119886 119887) lt 119862

1205733(119886 119887) 119860

1minus1205733(119886 119887)

(48)

1198621205724(119886 119887) 119860

1minus1205724(119886 119887)

lt 119873119860119862 (119886 119887) lt 119862

1205734(119886 119887) 119860

1minus1205734(119886 119887)

(49)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge 1

1205722le 23 120573

2ge 1 120572

3le 13 120573

3ge [2 log(2radic3 + log(2 + radic3)) minus

log 3](2 log 2)minus1 = 04648 1205724le 23 and120573

4ge [2 log(4120587+

3radic3) minus 2 log 6 minus log 3](2 log 2) = 07733

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin) 119902 isin (0 1205872)119903 isin (0 log(2 + radic3)) and 119904 isin (0 1205873) be the parameters suchthat 1 cosh(119901) = cos(119902) = 1 minus V2 cosh(119903) = sec(119904) = 1 + V2Then (9)ndash(12) lead to the conclusion that inequalities (46)ndash(49) are respectively equivalent to

1205721lt 1 minus

log119873119860119867 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119901 + sinh (119901) cosh (119901)) (2 sinh (119901) cosh (119901))]

log (1 cosh (119901))

= 1 minus 1205933(119901) lt 120573

1

(50)

1205722lt 1 minus

log119873119867119860 (119886 119887) minus log119860 (119886 119887)

log119867(119886 119887) minus log119860 (119886 119887)

= 1 minuslog [(119902 + sin (119902) cos (119902)) (2 sin (119902))]

log [cos (119902)]

= 1 minus 1205934(119902) lt 120573

2

(51)

1205723ltlog119873119862119860 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log ((119903 + sinh (119903) cosh (119903)) (2 sinh (119903)))

log [cosh (119903)]

= 1205935 (119903) lt 1205733

(52)

1205724ltlog119873119860119862 (119886 119887) minus log119860 (119886 119887)

log119862 (119886 119887) minus log119860 (119886 119887)

=log [(119904 + sin (119904) cos (119904)) (2 sin (119904) cos (119904))]

log [sec (119904)]

= 1205936 (119904) lt 1205734

(53)

Therefore Theorem 9 follows easily from (50)ndash(53) andLemmas 5ndash8

Theorem 10 Let 1205725 1205735 1205726 1205736isin [0 12] Then the double

inequalities

119867[1205725119886 + (1 minus 120572

5) 119887 1205725119887 + (1 minus 120572

5) 119886]

lt 119873119860119867 (119886 119887) lt 119867 [120573

5119886 + (1 minus 120573

5) 119887 1205735119887 + (1 minus 120573

5) 119886]

119867 [1205726119886 + (1 minus 120572

6) 119887 1205726119887 + (1 minus 120572

6) 119886]

lt 119873119867119860 (119886 119887) lt 119867 [120573

6119886 + (1 minus 120573

6) 119887 1205736119887 + (1 minus 120573

6) 119886]

(54)

hold for all 119886 119887 gt 0with 119886 = 119887 if and only if 1205725le 12minusradic66 =

009175 1205735ge 12 minusradic24 = 01464 120572

6le 12 minusradic36 =

02113 and 1205736ge 12 minus radic1 minus 12058742 = 02683

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120582 isin [0 12] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119901 isin (0infin)and 119902 isin (0 1205872) be the parameters such that 1 cosh(119901) =cos(119902) = 1 minus V2 Then from (9) and (10) we have

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119860119867 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [1 +

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887)

2

times [1 minus2(1 minus 2120582)

2(cosh (119901) minus 1)

cosh (119901)minus

119901

sinh (119901) cosh (119901)]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)

times [1 +sinh (119901) minus 119901

sinh (119901) (cosh (119901) minus 1)minus 2(1 minus 2120582)

2]

=119860 (119886 119887) [cosh (119901) minus 1]

2 cosh (119901)[1 + 120593

1(119901) minus 2(1 minus 2120582)

2]

(55)

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

119867[120582119886 + (1 minus 120582) 119887 120582119887 + (1 minus 120582) 119886] minus 119873119867119860 (119886 119887)

= 119860 (119886 119887) [1 minus (1 minus 2120582)2V2]

minus1

2119860 (119886 119887) [cos (119902) +

119902

sin (119902)]

=1

2119860 (119886 119887)

times [2 minus 2(1 minus 2120582)2(1 minus cos (119902)) minus cos (119902) minus

119902

sin (119902)]

=1 minus cos (119902)

2119860 (119886 119887) [1 minus 2(1 minus 2120582)

2minus 1205932(119902)]

(56)

Therefore Theorem 10 follows easily from (55) and (56)together with Lemmas 3 and 4

Theorem 11 Let 1205727 1205737 1205728 1205738isin [12 1] Then the double

inequalities

119862 [1205727119886 + (1 minus 120572

7) 119887 1205727119887 + (1 minus 120572

7) 119886]

lt 119873119862119860 (119886 119887) lt 119862 [1205737119886 + (1 minus 1205737) 119887 1205737119887 + (1 minus 1205737) 119886]

119862 [1205728119886 + (1 minus 120572

8) 119887 1205728119887 + (1 minus 120572

8) 119886]

lt 119873119860119862 (119886 119887) lt 119867 [120573

8119886 + (1 minus 120573

8) 119887 1205738119887 + (1 minus 120573

8) 119886]

(57)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205727le 12 +

radic36 = 07886 1205737ge 12 + radic6radic3 log(2 + radic3)12 =

08082 1205728le 12 + radic66 = 09082 and 120573

6ge 12 +

radic8radic3120587 minus 1812 = 09210

Proof Without loss of generality we assume that 119886 gt 119887 gt 0Let 120583 isin [12 1] V = (119886 minus 119887)(119886 + 119887) isin (0 1) 119903 isin (0 log(2 +radic3)) and 119904 isin (0 1205873) be the parameters such that cosh(119903) =sec(119904) = 1 + V2 Then from (11) and (12) one has

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119862119860 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [cosh (119903) + 119903

sinh (119903)]

=1

2119860 (119886 119887)

times [2 + 2(2120583 minus 1)2(cosh (119903) minus 1) minus cosh (119903) minus 119903

sinh (119903)]

=1

2119860 (119886 119887) [cosh (119903) minus 1] [2(2120583 minus 1)2 minus 1 + 1205931 (119903)]

119862 [120583119886 + (1 minus 120583) 119887 120583119887 + (1 minus 120583) 119886] minus 119873119860119862 (119886 119887)

= 119860 (119886 119887) [1 + (2120583 minus 1)2V2]

minus1

2119860 (119886 119887) [1 +

119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887)

times [1 + 2(2120583 minus 1)2(sec (119904) minus 1) minus 119904

sin (119904) cos (119904)]

=1

2119860 (119886 119887) [sec (119904) minus 1] [2(2120583 minus 1)2 minus 1 minus 1205932 (119904)]

(58)

where the functions1205931and1205932are defined as in Lemmas 3 and

4 respectivelyNote that

1205931[log (2 + radic3)] = 1 minus

radic3 log (2 + radic3)3

1205932(120587

3) =

4radic3120587 minus 18

9

(59)

Therefore Theorem 11 follows easily from Lemmas 3 and4 together with (58)-(59)

Theorem 12 Let 119878119860119867(119886 119887) = 119878[119860(119886 119887)119867(119886 119887)] 119878

119867119860(119886 119887) =

119878[119867(119886 119887) 119860(119886 119887)] 119878119860119862(119886 119887) = 119878[119860(119886 119887) 119862(119886 119887)] and

119878119862119860(119886 119887) = 119878[119862(119886 119887) 119860(119886 119887)] Then the inequalities

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119873119860119867 (119886 119887) lt 119878119867119860 (119886 119887)

lt 119873119867119860 (119886 119887) lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887)

lt 119878119862119860 (119886 119887) lt 119873119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119873119860119862 (119886 119887) lt 119862 (119886 119887)

(60)

hold for all 119886 119887 gt 0 with 119886 = 119887

Proof It follows from [1 (310)] and [11 (41)] together with[9 (5) Theorems 2 and 5] that

119873(119886 119887) gt119886 + 2119887

3gt SB (119886 119887) (61)

119867(119886 119887) lt 119878119860119867 (119886 119887) lt 119871 (119886 119887) lt 119878119867119860 (119886 119887) lt 119875 (119886 119887)

lt 119860 (119886 119887) lt 119872 (119886 119887) lt 119879 (119886 119887) lt 119878119862119860 (119886 119887)

lt 119876 (119886 119887) lt 119878119860119862 (119886 119887) lt 119862 (119886 119887)

(62)

Therefore the second fourth ninth and twelfth inequal-ities follow from (61) and the first sixth seventh eighth andeleventh inequalities follow from (62) immediately while thefifth and thirteenth inequalities can be derived from119867(119886 119887) lt

119860(119886 119887) lt 119862(119886 119887) and the fact that 119873119860119867(119886 119887) and 119873

119860119862(119886 119887)

are respectively the mean values of 119860(119886 119887) 119867(119886 119887) and119860(119886 119887) 119862(119886 119887)

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

Next we prove the third and tenth inequalities In factthe third inequality can be derived from the followinginequalities (63) [14 Theorem 12] and (64) [9 Theorem 3]together with 119860(119886 119887) gt 119867(119886 119887) Consider the following

119873119860119867 (119886 119887) lt

1

2119860 (119886 119887) +

1

2119867 (119886 119887) (63)

119878119867119860 (119886 119887) gt

2

120587119860 (119886 119887) + (1 minus

2

120587)119867 (119886 119887) (64)

while the tenth inequality can be derived from the inequality119873119862119860(119886 119887) lt 119862

1205733(119886 119887)119860

1minus1205733(119886 119887) in Theorem 9 and 119862(119886 119887) gt

119860(119886 119887) together with 119876(119886 119887) = 11986012(119886 119887)119862

12(119886 119887) where

1205733= [2 log(2radic3 + log(2 + radic3)) minus log 3](2 log 2) minus 1 =

04648

Remark 13 He et al [14 Theorem 12] Xia and Chu [17Theorem 31] and Chu et al [18 Theorem 21] proved thatthe double inequalities

1205721119860 (119886 119887) + (1 minus 1205721)119867 (119886 119887)

lt 119873119860119867 (119886 119887) lt 1205731119860 (119886 119887) + (1 minus 1205731)119867 (119886 119887)

1205722119860 (119886 119887) + (1 minus 1205722)119867 (119886 119887)

lt 119873119867119860 (119886 119887) lt 1205732119860 (119886 119887) + (1 minus 1205732)119867 (119886 119887)

1205723119860 (119886 119887) + (1 minus 1205723)119867 (119886 119887)

lt 119871 (119886 119887) lt 1205733119860 (119886 119887) + (1 minus 1205733)119867 (119886 119887)

1205724119860 (119886 119887) + (1 minus 1205724)119867 (119886 119887)

lt 119875 (119886 119887) lt 1205734119860 (119886 119887) + (1 minus 1205734)119867 (119886 119887)

(65)

hold for all 119886 119887 gt 0 with 119886 = 119887 if and only if 1205721le 13 120573

1ge

12 1205722le 23 120573

2ge 1205874 = 07853 120572

3le 0 120573

3ge 23

1205724le 2120587 = 06366 and 120573

4ge 56

From the above results we clearly see that the meanvalues 119873

119860119867(119886 119887) and 119871(119886 119887) and 119873

119867119860(119886 119887) and 119875(119886 119887) are

not comparable with each other

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This research was supported by the Natural Science Founda-tion of China under Grants 61374086 11371125 and 11171307and the Natural Science Foundation of Zhejiang Provinceunder Grant LY13A010004

References

[1] E Neuman and J Sandor ldquoOn the Schwab-Borchardt meanrdquoMathematica Pannonica vol 14 no 2 pp 253ndash266 2003

[2] E Neuman and J Sandor ldquoOn the Schwab-Borchardt mean IIrdquoMathematica Pannonica vol 17 no 1 pp 49ndash59 2006

[3] E Neuman ldquoInequalities for the Schwab-Borchardt mean andtheir applicationsrdquo Journal of Mathematical Inequalities vol 5no 4 pp 601ndash609 2011

[4] B C Carlson Special Functions of Applied Mathematics Aca-demic Press New York NY USA 1977

[5] B C Carlson ldquoAlgorithms involving arithmetic and geometricmeansrdquoThe American Mathematical Monthly vol 78 pp 496ndash505 1971

[6] J L Brenner and B C Carlson ldquoHomogeneous mean valuesweights and asymptoticsrdquo Journal of Mathematical Analysis andApplications vol 123 no 1 pp 265ndash280 1987

[7] Z-Y He Y-M Chu and M-K Wang ldquoOptimal bounds forNeuman means in terms of harmonic and contraharmonicmeansrdquo Journal of Applied Mathematics vol 2013 Article ID807623 4 pages 2013

[8] Y-M Chu and W-M Qian ldquoRefinements of bounds for Neu-manmeansrdquoAbstract and Applied Analysis vol 2014 Article ID354132 8 pages 2014

[9] E Neuman ldquoOn some means derived from the Schwab-Borchardt meanrdquo Journal of Mathematical Inequalities vol 8no 1 pp 171ndash183 2014

[10] E Neuman ldquoOn some means derived from the Schwab-Borchardt mean IIrdquo Journal of Mathematical Inequalities vol 8no 2 pp 361ndash370 2014

[11] E Neuman ldquoOn a new bivariatemeanrdquoAequationesMathemat-icae 2013

[12] Y Zhang Y-M Chu and Y-L Jiang ldquoSharp geometric meanbounds for Neuman meansrdquo Abstract and Applied Analysis vol2014 Article ID 949815 6 pages 2014

[13] Z-J Guo Y Zhang Y-M Chu and Y-Q Song ldquoSharp boundsfor Neuman means in terms of geometric arithmetic andquadratic meansrdquo In press httparxivorgabs14054384

[14] Z-Y He Y-M Chu Y-Q Song and X-J Tao ldquoNote oncertain inequalities for Neuman meansrdquo httparxivorgabs14054387

[15] G D Anderson S-L QiuM K Vamanamurthy andMVuori-nen ldquoGeneralized elliptic integrals and modular equationsrdquoPacific Journal of Mathematics vol 192 no 1 pp 1ndash37 2000

[16] S Simic and M Vuorinen ldquoLanden inequalities for zero-balanced hypergeometric functionsrdquo Abstract and AppliedAnalysis vol 2012 Article ID 932061 11 pages 2012

[17] W-F Xia and Y-M Chu ldquoOptimal inequalities related to thelogarithmic identric arithmetic and harmonic meansrdquo RevuedrsquoAnalyse Numerique et de Theorie de lrsquoApproximation vol 39no 4 pp 176ndash183 2010

[18] Y-M Chu Y-F Qiu M-K Wang and G-D Wang ldquoTheoptimal convex combination bounds of arithmetic and har-monic means for the Seiffertrsquos meanrdquo Journal of Inequalities andApplications vol 2010 Article ID 436457 7 pages 2010

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of