Download - 19 PP Chung Minh Bat Dang Thuc
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www.vnmath.com19 Phng php chng minh Bt ng thc
PHN 1CC KIN THC CN LU
1/nh ngha0
0
A B A B
A B A B
2/Tnh cht+ A>B AB B v B >C CA >+ A>B A+C >B + C+ A>B v C > D A+C > B + D+ A>B v C > 0 A.C > B.C+ A>B v C < 0 A.C < B.C+ 0 < A < B v 0 < C B > 0 A n > B n n+ A > B A n > B n vi n l+ A > B A n > B n vi n chn+ m > n > 0 v A > 1 A m > A n + m > n > 0 v 0 0 BA
11>
3/Mt s hng bt ng thc
+ A 2 0 vi A ( du = xy ra khi A = 0 )
+ An 0 vi A ( du = xy ra khi A = 0 )+ 0A vi A (du = xy ra khi A = 0 )
+ - A < A = A
+ A B A B+ + ( du = xy ra khi A.B > 0)
+ BABA ( du = xy ra khi A.B < 0)PHN II
CC PHNG PHP CHNG MINH BT NG THCPhng php 1 : Dng nh ngha
Kin thc : chng minh A > B. Ta lp hiu A B > 0Lu dng hng bt ng thc M 2 0 vi MV d 1 x, y, z chng minh rng :
a) x 2 + y 2 + z 2 xy+ yz + zxb) x 2 + y 2 + z 2 2xy 2xz + 2yz
c) x 2 + y 2 + z 2 +3 2 (x + y + z)Gii:
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www.vnmath.com19 Phng php chng minh Bt ng thc
a) Ta xt hiu : x 2 + y 2 + z 2 - xy yz zx =2
1.2 .( x 2 + y 2 + z 2 - xy yz
zx)
=2
10)()()( 222 ++ zyzxyx ng vi mi x;y;z R
V (x-y)2 0 vi
x ; y Du bng xy ra khi x=y(x-z)2 0 vix ; z Du bng xy ra khi x=z(y-z)2 0 vi z; y Du bng xy ra khi z=yVy x 2 + y 2 + z 2 xy+ yz + zx. Du bng xy ra khi x = y =z
b)Ta xt hiu: x 2 + y 2 + z 2 - ( 2xy 2xz +2yz ) = x 2 + y 2 + z 2 - 2xy +2xz 2yz
= ( x y + z) 2 0 ng vi mi x;y;z RVy x 2 + y 2 + z 2 2xy 2xz + 2yz ng vi mi x;y;z RDu bng xy ra khi x+y=zc) Ta xt hiu: x 2 + y 2 + z 2 +3 2( x+ y +z ) = x 2 - 2x + 1 + y 2 -2y +1 + z 2 -
2z +1= (x-1) 2 + (y-1) 2 +(z-1) 2 0. Du(=)xy ra khi x=y=z=1V d 2: chng minh rng :
a)222
22
+
+ baba; b)
2222
33
++
++ cbacbac) Hy tng qut bi
tonGii:
a) Ta xt hiu222
22
+
+ baba
= ( )4
2
4
2 2222 bababa ++
+= ( )abbaba 222
41 2222 + = ( ) 0
41 2 ba
Vy222
22
+
+ baba. Du bng xy ra khi a=b
b)Ta xt hiu
2222
33
++
++ cbacba= ( ) ( ) ( )[ ] 0
9
1 222 ++ accbba .Vy
2222
33
++++ cbacba
Du bng xy ra khi a = b =c
c)Tng qut2
2122
221 ........
+++
+++n
aaa
n
aaa nn
Tm li cc bc chng minh A B theo nh nghaBc 1: Ta xt hiu H = A - BBc 2:Bin i H=(C+D) 2 hoc H=(C+D) 2 +.+(E+F) 2
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www.vnmath.com19 Phng php chng minh Bt ng thc
Bc 3:Kt lun A BV d 1: Chng minh m,n,p,q ta u c : m 2 + n 2 + p 2 + q 2 +1 m(n+p+q+1) Gii:
014444
22
22
22
2
++
++
++
+ m
mqmq
mpmp
mnmn
m
012222
2222
+
+
+
mqmpmnm (lun ng)
Du bng xy ra khi
=
=
=
=
012
02
02
02
m
qm
pm
nm
=
=
=
=
22
2
2
m
mq
mp
mn
====
1
2
qpn
m
V d 2:Chng minh rng vi mi a, b, c ta lun c : )(444 cbaabccba ++++
Gii: Ta c : )(444 cbaabccba ++++ , 0,, > cba
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) 00)2(
)2()2(
0222
222
0222222
0
222222222222
22222
2222222222222222222
222
222222222222222
222444
222444
+++++
++
++++++
+++++
++
++
acabacbcbcabaccbba
abaacba
abcaccbacbcbbaaccbba
abcacbbca
caaccbcbbaba
abcacbbcacba
abcacbbcacba
ng vi mi a, b, c.Phng php 2 : Dng php bin i tng ng
Kin thc:Ta bin i bt ng thc cn chng minh tng ng vi bt ng thc
ng hoc bt ng thc c chng minh l ng.Nu A < B C < D , vi C < D l mt bt ng thc hin nhin, hoc bit l ng
th c bt ng thc A < B .Ch cc hng ng thc sau: ( ) 222 2 BABABA ++=+
( ) BCACABCBACBA 2222222
+++++=++ ( ) 32233 33 BABBAABA +++=+ V d 1: Cho a, b, c, d,e l cc s thc chng minh rng
a) abb
a +4
22
b) baabba ++++ 122
c) ( )edcbaedcba +++++++ 22222
Gii:
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www.vnmath.com19 Phng php chng minh Bt ng thc
a) abb
a +4
22 abba 44 22 + 044 22 + baa ( ) 02 2 ba
(BT ny lun ng). Vy abb
a +4
22 (du bng xy ra khi 2a=b)
b) baabba ++++ 122 ) )(21(2 22 baabba ++>++
0121222222
+++++ bbaababa 0)1()1()( 222 ++ baba Bt ng thc cui ng.Vy baabba ++++ 122 . Du bng xy ra khi a=b=1c) ( )edcbaedcba +++++++ 22222
( ) ( )edcbaedcba +++++++ 44 22222 ( ) ( ) ( ) ( ) 044444444 22222222 +++++++ cacadadacacababa
( ) ( ) ( ) ( ) 02222 2222 +++ cadacabaBt ng thc ng vy ta c iu phi chng minh
V d 2: Chng minh rng: ( )( ) ( )( )4488221010 babababa ++++Gii:( )( ) ( )( )4488221010 babababa ++++
128448121210221012 bbabaabbabaa ++++++ ( ) ( ) 022822228 + abbababa a2b2(a2-b2)(a6-b6) 0
a2b2(a2-b2)2(a4+ a2b2+b4) 0Bt ng thccui ng vy ta c iu phi chng minh
V d 3: cho x.y =1 v x y Chng minhyx
yx
+ 22
22
Gii: yx
yx
+ 22
22 v :xy nn x- y
0
x
2
+y
2 22 ( x-y)
x2+y2- 22 x+ 22 y 0 x2+y2+2- 22 x+ 22 y -2 0 x2+y2+( 2 )2- 22 x+ 22 y -2xy 0 v x.y=1 nn 2.x.y=2 (x-y- 2 )2 0 iu ny lun lun ng . Vy ta c iu phi chng
minhV d 4: Chng minh rng:a/ P(x,y)= 01269 222 ++ yxyyyx Ryx ,
b/ cbacba ++++ 222 (gi :bnh phng 2 v)c/ Cho ba s thc khc khng x, y, z tha mn:
++
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www.vnmath.com19 Phng php chng minh Bt ng thc
=(xyz-1)+(x+y+z)-xyz(zyx
111++ )=x+y+z - ( 0)
111>++
zyx(v
zyx
111++ < x+y+z
theo gt) 2 trong 3 s x-1 , y-1 , z-1 m hoc c ba s-1 , y-1, z-1 l dng.
Nu trng hp sau xy ra th x, y, z >1 x.y.z>1 Mu thun gt x.y.z=1 bt
buc phi xy ra trng hp trn tc l c ng 1 trong ba s x ,y ,z l s ln hn 1V d 5: Chng minh rng : 21
++++
, )3(cba
c
ca
c
++>+
Cng v theo v cc bt ng thc (1), (2), (3), ta c :
1>+++++ ca
c
cb
b
ba
a(*)
Ta c : )4(cba
ca
ba
abaa
+++b>c>0 v 1222 =++ cba . Chng minh rng3 3 3 1
2
a b c
b c a c a b+ +
+ + +Gii:
Do a,b,c i xng ,gi s ab c
+
+
+
ba
c
ca
b
cb
a cba222
p dng BT Tr- b-sp ta c
++
++
+++
+
++
++ ba
c
ca
b
cb
acba
ba
cc
ca
bb
cb
aa .
3...
222222 =
2
3.
3
1=
2
1
Vy2
1333
++
++
+ bac
ca
b
cb
aDu bng xy ra khi a=b=c=
3
1
V d 4: Cho a,b,c,d>0 v abcd =1 .Chng minh rng :( ) ( ) ( ) 102222 +++++++++ acddcbcbadcba
Gii: Ta c abba 222 + cddc 222 +
Do abcd =1 nn cd =ab
1(dng
2
11+
xx )
Ta c 4)1
(2)(2222 +=+++ab
abcdabcba (1)
Mt khc: ( ) ( ) ( )acddcbcba +++++ = (ab+cd)+(ac+bd)+(bc+ad)
= 222111
++
++
++
+
bcbc
acac
abab
Vy ( ) ( ) ( ) 102222 +++++++++ acddcbcbadcba
Phng php7 Bt ng thc Bernouli Kin thc:
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www.vnmath.com19 Phng php chng minh Bt ng thc
a)Dng nguyn thy: Cho a -1, n1 Z th ( ) naa n ++ 11 . Du = xy ra
khi v ch khi
=
=
1
0
n
a
b) Dng m rng:- Cho a > -1, 1 th ( ) naa ++ 11 . Du bng xy ra khi v ch khi a = 0.
- cho 10,1 + baba ab .Gii- Nu 1a hay 1b th BT lun ng- Nu 0 < a,b < 1
p dng BT Bernouli:( )11 1
1 1 .b b
bb aa a b aaa a a a a b
+ = + < + < > + Chng minh tng t:
ba
bb a
+> . Suy ra 1>+ ab ba (pcm).
V d 2: Cho a,b,c > 0.Chng minh rng5555
33
++++ cbacba . (1)
Gii
( ) 3333
1
555
+++
+++
++
cba
c
cba
b
cba
a
p dng BT Bernouli:
( )cba
acbcbaacb
cbaa
++++
++++=
++251213
55
(2)
Chng minh tng t ta uc:
( )
cba
bac
cba
b
+++
+
++
251
35
(3)
( )
cba
cba
cba
c
+++
+
++
251
35
(4)
Cng (2) (3) (4) v theo v ta c
+++ +++ ++3333
555
cbac
cbab
cbaa (pcm)
Ch : ta c bi ton tng qut sau y:Cho .1;0,..., 21 > raaa n Chng minh rng
r
nrn
rr
n
aaa
n
aaa
++++++ ........ 2121 .
Du = naaa === ....21 .(chng minh tng t bi trn).V d 3:Cho 1,,0 zyx . Chng minh rng
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www.vnmath.com19 Phng php chng minh Bt ng thc
( )( )8
81222222 ++++ zyxzyx .
Giit ( )2,,12,2,2 === cbacba zyx .
( )( )
)1(32
023
02121
2
++
aaaa
aaa
Chng minh tng t:
)3(32
)2(32
+
+
cc
bb
Cng (1) (2) (3) v theo v ta c
( ) ( )
)(111
)(8
81
11122
11129
pcmcbacba
cbacba
cbacba
csi
++++
++++
+++++
Ch : Bi ton tng qut dng ny Cho n s [ ] 1,,,....,, 21 > cbaxxx nTa lun c:
( )( ) ( )[ ]ba
baxxxxxx
c
ccncccccc nn
+
+++++++
4........
2
2121
Phng php 8: S dng tnh cht bc cuKin thc: A>B v B>C th A>C
V d 1: Cho a, b, c ,d >0 tha mn a> c+d , b>c+dChng minh rng ab >ad+bc
Gii:
Tac
+>+>
dcb
dca
>>>>
0
0
cdb
dca (a-c)(b-d) > cd
ab-ad-bc+cd >cd ab> ad+bc (iu phi chng minh)
V d 2: Cho a,b,c>0 tha mn3
5222 =++ cba . Chng minh
abccba
1111
0 ta c
cba
111+
abc
1
V d 3: Cho 0 < a,b,c,d 1-a-b-c-d Gii:Ta c (1-a).(1-b) = 1-a-b+ab
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www.vnmath.com19 Phng php chng minh Bt ng thc
Do a>0 , b>0 nn ab>0 (1-a).(1-b) > 1-a-b (1)Do c 0 ta c (1-a).(1-b) ( 1-c) > 1-a-b-c
(1-a).(1-b) ( 1-c).(1-d) > (1-a-b-c) (1-d) =1-a-b-c-d+ad+bd+cd (1-a).(1-b) ( 1-c).(1-d) > 1-a-b-c-d (iu phi chng minh)
V d 4: Cho 0 b
ath
cb
ca
b
a
++
>
b Nu 1+++>
>
..
222222
222222222
V d2 (HS t gii)1/ Cho a,b,c l chiu di ba cnh ca tam gic
Chng minh rng )(2222
cabcabcbacabcab ++
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x + y = 1
MN
O
MKH
M
x
y
www.vnmath.com19 Phng php chng minh Bt ng thc
Th bu = , av = ; )()(. cbccacvu +=Hn na: += abcbccacvuvuvuvu )()(.),cos(... (PCM)V d 2:
Cho 2n s: niyx ii ,...,2,1,; = tha mn: .111
=+ ==
n
i
i
n
i
i yx Chng minh rng:
22
1
22 +=n
i
ii yx
Gii:V hnh
Trong mt phng ta , xt:),( 111 yxM : ),( 21212 yyxxM ++ ;; ),( 11 nnn yyxxM ++++
Gi thit suy ra n
M ng thng x + y = 1. Lc :2
1
2
11 yxOM += ,2
2
2
221 yxMM += ,2
3
2
332 yxMM += ,,22
1 nnnn yxMM +=
V 1OM 21MM+ 32MM+2
21 =++ OHOMMM nnn
+ = 2
2
1
22n
i
iiyx (PCM)
Phng php 13: i bin s
V d1: Cho a,b,c > 0 Chng minh rng2
3
++
++
+ bac
ac
b
cb
a(1)
Gii: t x=b+c ; y=c+a ;z= a+b ta c a=2
xzy +; b =
2
yxz +; c =
2zyx +
ta c (1) z
zyx
y
yxz
x
xzy
222
++
++
+
2
3
3111 +++++z
y
z
x
y
z
y
x
x
z
x
y ( 6)()() +++++
z
y
y
z
z
x
x
z
y
x
x
y
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www.vnmath.com19 Phng php chng minh Bt ng thc
Bt ng thc cui cng ng v ( ;2+y
x
x
y 2+
z
x
x
z; 2+
z
y
y
znn ta c
iu phi chng minh V d2:
Cho a,b,c > 0 v a+b+c 0 , b > 0 , c > 0 CMR: 81625
>+
++
++ ba
c
ac
b
cb
a
2)Tng qut m, n, p, q, a, b >0CMR
( ) ( )pnmpnmba
pc
ac
nb
cb
ma++++
++
++
+2
2
1
Phng php 14: Dng tam thc bc hai
Kin th: Cho f(x) = ax2 + bx + cnh l 1:
f(x) > 0,
0
0ax
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www.vnmath.com19 Phng php chng minh Bt ng thc
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www.vnmath.com19 Phng php chng minh Bt ng thc
Gii: Bt ng thc cn chng minh tng ng vi ( ) 044.22 322242 >++++ xyxxyyxyx ( ) 0414.)1( 22222 >+++ yxyyxy
Ta c ( ) ( ) 0161414 2222222 +y vy ( ) 0, >yxf (pcm)
Phng php 15: Dng quy np ton hcKin thc: chng minh bt ng thc ng vi 0nn > ta thc hin cc bc sau :1 Kim tra bt ng thc ng vi 0nn =2 - Gi s BT ng vi n =k (thay n =k vo BT cn chng minh c
gi l gi thit quy np )3- Ta chng minh bt ng thc ng vi n = k +1 (thay n = k+1vo BT
cn chng minh ri bin i dng gi thit quy np)4 kt lun BT ng vi mi 0nn >
V d1: Chng minh rng : nn
1
2
1
....2
1
1
1222
nNn
(1)
Gii: Vi n =2 ta c2
12
4
11
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www.vnmath.com19 Phng php chng minh Bt ng thc
V tri (2) 242
.2
1111 ++++ +
+++=
++ kkkkkkkk babbaabababa
042
1111
+++
+ ++++ kkkkkk bbaababa
( ) ( ) 0. baba kk (3)Ta chng minh (3)
(+) Gi s a b v gi thit cho a -b a b k
kk bba ( ) ( ) 0. baba kk
(+) Gi s a < b v theo gi thit - a
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www.vnmath.com19 Phng php chng minh Bt ng thc
n=k ( k ):gi s bt ng thc ng, tc l:))(()( 222
2
1
22
2
2
1
2
2211 kkkk bbbaaabababa +++++++++
n= k+1 . Ta cn chng minh:))(()( 2 1
22
21
21
22
21
2112211 ++++ +++++++++ kkkk bbbaaabababa (1)
Tht vy: 222122
2
2
1
22
2
2
1 ).())(()1( baabbbaaaVP kkk +++++++++= +2
1
2
1
22
2
2
1
2
.)( +++++++ kkk babbba ++++++ ++++ 112211112211 22)( kkkkkk bababababababa
2
1
2
1112 ++++ +++ kkkkkk bababa
2)( 22211 ++++ kkbababa )( 2211 kkbababa +++ 11 ++ kk ba2
1
2
1.++
+ kk ba2
112211)( +++++ kk bababa
Vy (1) c chng minhV d 6: Cho n1 , niRba ii ,...,2,1,, = . Chng minh rng:
n
aaa
n
aaa nn22
2
2
1221 )(+++
+++
Gii:n=1: Bt ng thc lun ngn=k ( k ):gi s bt ng thc ng, tc l:
k
aaa
k
aaa kk22
2
2
1221 )(+++
+++
n= k+1 . Ta cn chng minh:1
)1
(2
1
2
2
2
12121
++++
+
+++ ++k
aaa
k
aaa kk (1)
t:k
aaaa k 132 +
+++=
)2(1
1)1(
1
222
1akaaka
kVP
+++=
+++++
++++
+ ++
k
aaakak
k
aaaka
k
kk
2
1
2
3
2
22
1
2
1
2
3
2
222
12.
)1(
1
1
2
1
2
2
2
1
++++
= +k
aaa k
Vy (1) c chng minh
V d 7: Chng minh rng: 2,,)1( 1 +> nnnn nn
Gii: n=2
=+
=
3)1(
4
1n
n
n
n 1)1( +> nn nn
n=k 2 : gi s bt ng thc ng, tc l: 1)1( +> kk kkn= k+1:Ta c : 111 )1()1()1( ++ +++ kkkk kkkk
212222)1(])1[()1()1( ++=++= kkkk kk
)2()2(212 kkkk k ++> (v kkkkk 212)1( 222 +>++=+ )
kkkk )2( + kk kk )2()1( 1 +>+ + Bt ng thc ng vi n= k+1
Vy 2,,)1( 1 +> nnnn nn
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www.vnmath.com19 Phng php chng minh Bt ng thc
V d 8: Chng minh rng: Rxnxnnx ,,sinsinGii: n=1: Bt ng thc lun ngn=k :gi s bt ng thc ng, tc l: xkkx sinsin n= k+1 . Ta cn chng minh: xkxk sin)1()1sin( ++
Ta c:
++
Rxxx
Rbababa
,1c o s,s i n
,,
Nn: xkxxkxxk sincoscossin)1sin( +=+xkxxkx sin.coscos.sin + xkx sin..sin + xxk sin..sin + xk sin)1( +=
Bt ng thc ng vi n= k+1. Vy: Rxnxnnx ,,sinsin + Phng php 16: Chng minh phn chngKin thc:1) Gi s phi chng minh bt ng thc no ng , ta hy gi s bt
ng thc sai v kt hp vi cc gi thit suy ra iu v l , iu v l c thl iu tri vi gi thit , c th l iu tri ngc nhau .T suy ra bt ng thccn chng minh l ng
2) Gi s ta phi chng minh lun p qMun chng minh qp (vi p : gi thit ng, q : kt lun ng) php
chng minh c thc hin nh sau:Gi s khng c q ( hoc q sai) suy ra iu v l hoc p sai. Vy phi c
q (hay q ng)Nh vy ph nh lun ta ghp tt c gi thit ca lun vi ph nh
kt lun ca n .
Ta thng dng 5 hnh thc chng minh phn chng sau :A - Dng mnh phn o : P QB Ph nh ri suy tri gi thitC Ph nh ri suy tri vi iu ngD Ph nh ri suy ra 2 iu tri ngc nhauE Ph nh ri suy ra kt lun :
V d 1: Cho ba s a,b,c tha mn a +b+c > 0 , ab+bc+ac > 0 , abc > 0Chng minh rng a > 0 , b > 0 , c > 0
Gii:Gi s a 0 th t abc > 0 a 0 do a < 0. M abc > 0 v a < 0 cb 0 a(b+c) > -bc > 0V a < 0 m a(b +c) > 0 b + c < 0
a < 0 v b +c < 0 a + b +c < 0 tri gi thit a+b+c > 0Vy a > 0 tng t ta c b > 0 , c > 0V d 2:Cho 4 s a , b , c ,d tha mn iu kin
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www.vnmath.com19 Phng php chng minh Bt ng thc
ac 2.(b+d) .Chng minh rng c t nht mt trong cc bt ng thc sau lsai: ba 42 < , dc 42 0Trong ba s x-1 , y-1 , z-1 ch c mt s dngTht vy nu c ba s dng th x,y,z > 1 xyz > 1 (tri gi thit)Cn nu 2 trong 3 s dng th (x-1).(y-1).(z-1) < 0 (v l)Vy c mt v ch mt trong ba s x , y,z ln hn 1V d 4: Cho 0,, >cba v a.b.c=1. Chng minh rng: 3++ cba (Bt ng
thc Cauchy 3 s)
Gii: Gi s ngc l i:3
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www.vnmath.com19 Phng php chng minh Bt ng thc
acb < 22)( bac > 0))(( >+++ cbacba (2)bac < 22)( cba > 0))(( >+++ cbacba (3)
Nhn (1), (2) v (3) v vi v ta c:0)])()([(
2 >++++ cbacbacba V l. Vy bi ton c chng minh
Phng php 17 : S dng bin i lng gic1. Nu Rx th t x = Rcos , [ ] ,0 ; hoc x = Rsin
2,
2,
2. Nu Rx th t x =cos
R [ )
2
3,,0
c
3.Nu ( ) ( ) )0(,222 >=+ Rbyax th t )2(,
s i n
c o s
=
+=
+=
Rby
Rax
4. Nu 0,222
>=
+
baRb
y
a
x th t )2(,s i n
c o s
=
+=
+=
b Ry
a Rx
5. Nu trong bi ton xut hin biu thc : ( ) ( )0,,22 >+ babax
Th t:
=
2,
2,
tg
a
bx
V d 1: Cmr : ( )( )( ) [ ]1,1,,211311 2222 ++ baabababbaGii : 1,1 ba
t :
=
=
c o s
c o
b
a [ ]( ) ,0,
Khi :
( ) ( )( )( )
[ ]
2 2 2 21 1 3 1 1
cos .sin cos .sin 3 cos .cos sin .sin
sin( ) 3.cos( ) 2cos( ) 2,2 ( )6
a b b a ab b a
dpcm
+ +
= + +
= + + + = +
V d 2 : Cho 1, ba .Chng minh rng : ababba + 11Gii :
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www.vnmath.com19 Phng php chng minh Bt ng thc
t :
=
=
2,0,
cos
1co s
1
2
2
b
a
22 2
2 2 2 2 2 2
2 2 2 2 2 2
1 1 ( .cos .co1 1
cos cos cos cos cos .cos1 (sin 2 sin 2 ) sin( )cos ( ) 1
2 cos .cos cos .cos cos .cos
tg tg tg tg a b b a tg tg
inab
+ + = + = + =
+ + = = =
V d 3: Cho 0ab .Chng minh rng : 2224
)4(222
22
22
+
ba
baa
Gii :t:
=
22,
2,2
btga
2 2 2 22
2 2 2
( 4 ) ( 2)4( 1).cos
4 1
2sin 2 2(1 cos 2 ) 2(sin 2 cos 2 ) 22 2 sin(2 ) 2 2 2 2, 2 2 2
2
a a b tg tg tg
a b tg
= =
+ +
= + = =
Phng php 18: S dng khai trin nh thc Newton.Kin thc:Cng thc nh thc Newton
( ) RbaNnbaCban
k
kknkn
n + =
,,, *
0
.
Trong h s )0(!)!(
!nk
kkn
nC
k
n
= .
Mt s tnh cht t bit ca khai trin nh thc Newton:+ Trong khai trin (a + b)n c n + 1 s hng.+ S m ca a gim dn t n n 0, trong khi s m ca b tng t 0 n n.
Trong mi s hng ca khai trtin nh thc Newton c tng s m ca a v b bngn.
+Cc h s cch u hai u th bng nhau knn
kn CC
= .+ S hng th k + 1 l )0(. nkbaC kknkn
V d 1:
Chng minh rng ( )*,0,11 Nnanaa n ++ (bt ng thc bernoulli)
Gii
Ta c: ( ) naaCCaCa nnn
k
kk
n
n +=+=+ =
11 10
0
(pcm)
V d 2:Chng minh rng:
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www.vnmath.com19 Phng php chng minh Bt ng thc
a) *,0,,22
Nnbababa
nnn
+
+
b) *,0,,,33
Nncbacbacba
nnnn
++
++
Gii
Theo cng thc khai trin nh thc Newton ta c:( )( )
( )
( )( )( )
n
baba
baCCCCba
abCbaCbaCbaCba
bababababa
niba
abCabbaCabbaCbaCba
aCabCabCbCba
bCbaCbaCaCba
nnn
nnnnn
nnnn
nn
nnnn
nnnn
nnn
nnn
n
iniiinnniiinin
nnnn
nnnn
nnn
nnn
n
nnn
nnn
nn
nn
n
nnn
nnn
nn
nn
n
+
+
+=+++++=
+++++++++
+
=
++++++++=+
++++=+
++++=+
2
)(2)....)((
)()(....)()(2
0
:1,...,2,1,0,
)()..(....)()(2
.....
.....
110
110
1111110
11110
11110
b) t 03
++= cbad
Theo cu (a) ta c:
n
nnnn
nnnnnnnnn
nn
nn
nn
nnnn
cbad
cba
dcbaddcba
ddcba
dcba
dcba
dcba
++=
++
+++++
+++
++
+
=
++
+
+++
33
34
)4
(2
22
4
22
22
4
Phng php 19: S dng tch phnHm s: [ ] Rbagf ,:, lin tc, lc :
* Nu [ ]baxxf ,,0)( th b
a
dxxf 0)(
* Nu [ ]baxxgxf ,),()( th b
a
b
a
dxxgdxxf )()(
* Nu [ ]baxxgxf ,),()( v [ ] )()(:, 000 xgxfbax > th
b
a
b
a
dxxgdxxf )()( .
* b
a
b
a
dxxfdxxf )()( .
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www.vnmath.com19 Phng php chng minh Bt ng thc
* Nu [ ]baxMxfm ,,)( th b
a
Mdxxfab
m )(1
(m, M l hng
s)V d 1: Cho A, B, C l ba gc ca tam gic.
Chng minh rng: 3222
++C
tgB
tgA
tg
Gii:
t ),0(,2
)( = xx
tgxf
),0(,0)2
1(22
1)(
)2
1(2
1)(
2''
2'
>+=
+=
xx
tgx
tgxf
xtgxf
p dng bt ng thc Jensen cho:
3222
63
222
63
222
33
)()()(
++
++
++++
++
++
Ctg
Btg
Atg
tgC
tgB
tgA
tg
CBAtg
Ctg
Btg
Atg
CBAf
CfBfAf
V d 2: Chng minh:6cos2510
2
0
2
xdx
Gii
Trn on
2,0 ta c:
( )
2 2 2
22
2 2
2 20 0
0 cos 1 0 2cos 2 2 2cos 0
1 1 13 5 2cos 5
5 5 2cos 3
1 10 0
5 2 5 2cos 3 2 10 5 2cos 6
x x x
xx
dx dxpcm
x x
PHN III : CC BI TP NNG CAO
*Dng nh ngha1) Cho abc = 1 v 363 >a . . Chng minh rng +
3
2ab2+c2> ab+bc+ac
Gii: Ta xt hiu: +3
2ab2+c2- ab- bc ac = +
4
2a+
12
2ab2+c2- ab- bc ac
= ( +4
2ab2+c2- ab ac+ 2bc) +
12
2a3bc =(
2
a-b- c)2 +
a
abca
12
363
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www.vnmath.com19 Phng php chng minh Bt ng thc
=(2
a-b- c)2 +
a
abca
12
363 >0 (v abc=1 v a3 > 36 nn a >0 )
Vy : +3
2ab2+c2> ab+bc+ac iu phi chng minh
2) Chng minh rng
a) )1.(212244
+++++ zxxyxzyxb) vi mi s thc a , b, c ta c036245
22 >+++ baabbac) 024222 22 +++ baabba
Gii:a) Xt hiu: xxzxyxzyx 22221 222244 ++++ = ( ) ( ) ( ) 22222 1++ xzxyx
= HH 0 ta c iu phi chng minhb) V tri c th vit H = ( ) ( ) 1112 22 +++ bba H > 0 ta c pcm
c) v tri c th vit H = ( ) ( )22
11 ++ bba H 0 ta c iu phi chngminh
* Dng bin i tng ng1) Cho x > y v xy =1 .Chng minh rng
( )( )
82
222
+
yx
yx
Gii: Ta c ( ) ( ) 22 2222 +=+=+ yxxyyxyx (v xy = 1)
( ) ( ) ( ) 4.4 24222 ++=+ yxyxyxDo BT cn chng minh tng ng vi ( ) ( ) ( ) 224 .844 yxyxyx ++
( ) ( ) 044 24 + yxyx ( )[ ] 02 22 yxBT cui ng nn ta c iu phi chng minh
2) Cho xy 1 .Chng minh rng
xyyx +
++
+ 12
1
1
1
122
Gii:
Ta c xyyx ++++ 12
1
1
1
122 01
1
1
1
1
1
1
1222
+++
++ xyyyx
( ) ( ) ( ) ( ) 01.11.1 22
2
2
++
+
++
xyy
yxy
xyx
xxy ( ) ( ) ( ) ( ) 01.1)(
1.1
)(22
++
+
++
xyy
yxy
xyx
xyx
( ) ( )
( ) ( ) ( ) 01.1.11
22
2
+++
xyyx
xyxyBT cui ny ng do xy > 1 .Vy ta c pcm
* Dng bt ng thc ph
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www.vnmath.com19 Phng php chng minh Bt ng thc
1) Cho a , b, c l cc s thc v a + b +c =1 Chng minh rng3
1222 ++ cba
Gii: p dng BT BunhiaCpski cho 3 s (1,1,1) v (a,b,c)Ta c ( ) ( ) ( )2222 .111.1.1.1 cbacba ++++++ ( ) ( )2222 .3 cbacba ++++
3
1222 ++ cba (v a+b+c =1 ) (pcm)
2) Cho a,b,c l cc s dng . Chng minh rng ( ) 9111
.
++++
cbacba
(1)
Gii: (1) 9111 ++++++++a
c
a
c
c
b
a
b
c
a
b
a
93
++
++
++
b
c
c
b
a
c
c
a
a
b
b
a
p dng BT ph 2+ xy
y
x
Vi x,y > 0. Ta c BT cui cng lun ng
Vy ( ) 9111
.
++++
cbacba (pcm)
* Dng phng php bc cu1) Cho 0 < a, b,c
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www.vnmath.com19 Phng php chng minh Bt ng thc
d a d a d a c
a b c d d a b a b c d
+ + + +< 0 ,p dng BT Csi ta c
x+ y + z 33 xyz 31 1
3 27 xyz xyz
p dng bt ng thc Csi cho x+y ; y+z ; x+z ta c
( ) ( ) ( ) ( ) ( ) ( )3. . 3 . .x y y z z x x y y z x z+ + + + + + ( ) ( ) ( )32 3 . .x y y z z x + + +
Du bng xy ra khi x=y=z=1
3
Vy S 8 1 8
.27 27 729
= . Vy S c gi tr ln nht l8
729khi x=y=z=
1
3 V d 3: Cho xy+yz+zx = 1. Tm gi tr nh nht ca 4 4 4 x y z+ + Gii: p dng BT Bunhiacpski cho 6 s (x,y,z) ;(x,y,z)
Ta c ( ) ( ) 22 2 2 2 xy yz zx x y z+ + + + ( ) 22 2 21 x y z + + (1)
p dng BT Bunhiacpski cho (2 2 2
, ,x y z ) v (1,1,1)Ta c 2 2 2 2 2 2 2 4 4 4 2 2 2 2 4 4 4( ) (1 1 1 )( ) ( ) 3( )x y z x y z x y z x y z+ + + + + + + + + +
T (1) v (2) 4 4 41 3( ) x y z + + 4 4 41
3 x y z + +
Vy 4 4 4 x y z+ + c gi tr nh nht l1
3khi x=y=z= 3
3
V d 4: Trong tam gic vung c cng cnh huyn , tam gic vung no cdin tch ln nht Gii: Gi cnh huyn ca tam gic l 2a
ng cao thuc cnh huyn l hHnh chiu cc cnh gc vung ln cnh huyn l x,y
Ta c S = ( ) 21
. . . . .2
x y h a h a h a xy+ = = =
V a khng i m x+y = 2a. Vy S ln nht khi x.y ln nht x y =Vy trong cc tam gic c cng cnh huyn th tam gic vung cn c din tch
ln nht2/ Dng Bt ng thc gii phng trnh v h phng trnh
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www.vnmath.com19 Phng php chng minh Bt ng thc
V d 1:Gii phng trnh: 2 2 24 3 6 19 5 10 14 4 2 x x x x x x+ + + + + = Gii : Ta c 23 6 19x x+ + 23.( 2 1) 16x x= + + + 23.( 1) 16 16x= + +
( )225 10 14 5. 1 9 9 x x x+ + = + +
Vy 2 24. 3 6 19 5 10 14 2 3 5 x x x x+ + + + + + =
Du ( = ) xy ra khi x+1 = 0
x = -1Vy 2 2 24 3 6 19 5 10 14 4 2 x x x x x x+ + + + + = khi x = -1Vy phng trnh c nghim duy nht x = -1
V d 2: Gii phng trnh 2 22 4 4 3 x x y y+ = + + Gii : p dng BT BunhiaCpski ta c :
( )2 2 2 2 22 1 1 . 2 2. 2 2 x x x x+ + + = Du (=) xy ra khi x = 1
Mt khc ( )224 4 3 2 1 2 2 y y y+ + = + + Du (=) xy ra khi y = -
1
2
Vy2 2
2 4 4 3 2 x x y y+ = + + = khi x =1 v y =-
1
2
Vy nghim ca phng trnh l1
1
2
x
y
=
=
V d 3:Gii h phng trnh sau: 4 4 41 x y z
x y z xyz
+ + = + + =
Gii: p dng BT Csi ta c4 4 4 4 4 4
4 4 4 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2 2 2
x2 2 2
2 2 2
x y y z z x y z x y y z z x
x y y z z y z z x z y x
+ + ++ + = + + + +
+ + + + +
2 2 2 .( ) y xz z xy x yz xyz x y z + + + +
V x+y+z = 1) Nn 4 4 4 x y z xyz+ + Du (=) xy ra khi x = y = z =1
3
Vy 4 4 41 x y z
x y z xyz
+ + = + + =
c nghim x = y = z =1
3
V d 4: Gii h phng trnh sau2
2
4 8
2
xy y
xy x
=
= +
(1)
(2)
T phng trnh (1) 28 0y hay 8y
T phng trnh (2) 2 2 . 2 2 x x y x + =
2 2 22 2 2 0 ( 2) 0 2 2 x x x x x + = =
Nu x = 2 th y = 2 2Nu x = - 2 th y = -2 2
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www.vnmath.com19 Phng php chng minh Bt ng thc
Vy h phng trnh c nghim2
2
x
y
=
= v
2 2
2 2
x
y
=
= 3/ Dng BT gii phng trnh nghim nguyn
V d 1: Tm cc s nguyn x,y,z tho mn 2 2 2 3 2 3 x y z xy y z+ + + +
Gii: V x,y,z l cc s nguyn nn 2 2 2 3 2 3 x y z xy y z+ + + +
( )2 2
2 2 2 2 233 2 3 0 3 3 2 1 04 4
y y x y z xy y z x xy y z z
+ + + + + + + +
( )2 2
23 1 1 0
2 2
y yx z
+ +
(*)
M ( )2 2
23 1 1 0
2 2
y yx z
+ +
,x y R
( )2 2
23 1 1 02 2y yx z + + =
02 1
1 0 22
11 0
yx
xy
y
zz
==
= = = =
Cc s x,y,z phi tm l
1
21
x
yz
=
= =
V d 2: Tm nghim nguyn dng ca phng trnh1 1 1
2 x y z
+ + =
Gii: Khng mt tnh tng qut ta gi s x y z
Ta c1 1 1 3
2 2 3z x y z z
= + +
M z nguyn dng vy z = 1. Thay z = 1 vo phng trnh ta c1 1
1
x y
+ =
Theo gi s x y nn 1 =1 1
x y+
1
y 2y m y nguyn dng
Nn y = 1 hoc y = 2Vi y = 1 khng thch hpVi y = 2 ta c x = 2Vy (2 ,2,1) l mt nghim ca phng trnh
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Hon v cc s trn ta c cc nghim ca phng trnh l (2,2,1);(2,1,2);(1,2,2)
V d 3:Tm cc cp s nguyn tho mn phng trnh x x y+ = (*)Gii:
(*) Vi x < 0 , y < 0 th phng trnh khng c ngha
(*) Vi x > 0 , y > 0Ta c x x y+ = 2 x x y + = 2 0 x y x = >
t x k= (k nguyn dng v x nguyn dng )Ta c 2.( 1)k k y+ =
Nhng ( ) ( )22 1 1k k k k < + < + 1k y k < < +
M gia k v k+1 l hai s nguyn dng lin tip khng tn ti mt snguyn dng no c
Nn khng c cp s nguyn dng no tho mn phng trnh .
Vy phng trnh c nghim duy nht l :0
0
x
y
= =
Bi tp ngh :
Bi 1:Chng minh rng vi mi a,b,c > 0 :cbaab
c
ac
b
bc
a 111++++
HD : Chuyn v quy ng mu a v tng bnh phng cc ng thc.
Bi 2:Chng minh bt ng thc : *)(1)1(
1..
4.3
1
3.2
1
2.1
1Nn
nn 0 v a + b + c 1. Cmr : 641
1
1
1
1
1
+
+
+ cba
HD : p dng bt ng thc Csi cho
+
+
+
cba
11,
11,
11
Bi 4 : Cho 0,0 cbca . Cmr : abcbccac + )()(
HD : p dng bt ng thc Csi chob
cb
a
c
a
ca
b
c , , ri cng hai v
theo v.
Bi 5: Cho a, b >1. Tm GTNN ca S =11
22
+
ab
b
a
HD : p dng bt ng thc Csi cho 1,1
22
abba v xt trng hp du= xy ra .
Bi 9 : Tm GTLN v GTNN ca y = 2242
)21(
1283
x
xx
+++
HD: t x=
2,
2,
2
1 tg
Bi 10: Cho 36x .916 22 =+ y Cmr :4
2552
4
15+ xy
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HD: t :
=
=
sin4
3
cos2
1
y
x
Bi 11: Cmr : [ ]1,1),121(2
1122 ++ xx
xx
HD : t x =
4,
4,2sin
Bi 12: Cho 1,0, cba . Chng minh rng: accbbacba 222222 1 +++++Bi 13: Cho ABC c a, b, c l di cc cnh. Chng minh rng:
0)()()(222 ++ acaccbcbbaba
Bi 14: Cho 0,,1, bann . Chng minh rngnnn baba
++
22
Bi 15: nn 2, . Chng minh rng: 31
12