phuong phap chung minh bdt 1

7/23/2019 Phuong Phap Chung Minh Bdt 1

1/11

1

BA PHNG PHP CHNG MINH BT NG THC

TS. Phm ThBch NgcTp ch Ton hc v Tui tr

I. Sdng tnh cht tip tuyn ca hm s tng chnh ca phng php l sdng cng thc phng trnh tip tuynca mt thhm s tm mt biu thc trung gian trong cc nh gi btng thc.Tnh cht. Cho hm s ( )f x xc nh, lin tc v c o hm trn K. Khi

tip tuyn ti mt im 0x K c phng trnh ( ) ( ) ( )0 0 0'y f x x x f x= + thng

nm trn (hoc nm di) thhm s f trn K, nn ta c

( ) ( )( ) ( )0 0 0'f x f x x x f x + (hoc ( ) ( )( ) ( )0 0 0'f x f x x x f x + ) vi mi x K .

Ttnh cht ny, ta thy vi mi 1 2, ,..., nx x x K ta c

( ) ( ) ( ) ( ) ( ) ( )1 2 0 1 2 0 0... ' ...n nf x f x f x f x x x x nx nf x+ + + + + + +

hoc ( ) ( ) ( ) ( ) ( ) ( )1 2 0 1 2 0 0... ' ...n nf x f x f x f x x x x nx nf x+ + + + + + + .

Nhvy, nu mt bt ng thc c dng tng hm nhvtri ca bt ngthc trn, v c githit 1 2 0... nx x x nx+ + + = vi ng thc xy ra khi tt cccbin ix u bng nhau v bng 0x , th ta c thhi vng chng minh n bngphng php tip tuyn.V d1. (FRANCE 2007)

Cho a, b, c, d l cc sthc dng sao cho 1a b c d + + + = .

Chng minh rng ( ) ( )3 3 3 3 2 2 2 2 168

a b c d a b c d + + + + + + + .

Li gii

Tgithit suy ra ( ), , , 0;1a b c d . t ( ) 3 26f x x x= , vi ( )0;1x

Khi bt ng thc trthnh ( ) ( ) ( ) ( )1

8f a f b f c f d+ + + .

Ta don ng thc xy ra khi 14

a b c d = = = = . V vy ta tm phng

trnh tip tuyn ca thhm s ( )f x ti im1 1

;4 32

M

.

Phng trnh tip tuyn l 1 1 1 5 1'4 4 32 8

xy f x

= + =

.

Bng cch phc tho th hm s ta nhn thy tip tuyn ti M nmdi thhm s ( )y f x= trong khon (0 ; 1) nn nh hng chng

minh BT

( ) 3 2 3 25 1 5 1

, (0;1) 6 48 8 5 1 08 8

x xf x x x x x x x

+

( ) ( ) ( )2

4 1 3 1 0, 0;1x x x +


2/11

2

Do ( ) ( ) ( ) ( )5 5 5 5 4 1

8 8

a b c d f a f b f c f d

+ + + + + + =

ng thc xy ra khi 14

a b c d = = = = .(pcm)

V d2. (USA 2003) Cho , ,a b c l cc sthc dng. Chng minh rng

( )( )

( )( )

( )( )

82

22

22

2 22

2

22

2

22

2

++

+++++

+++++

++bac

bacacb

acbcba

cba .

Li giiBT c tnh thun nht. Khng mt tng qut, ta c th gi s rng

1a b c+ + = . Khi BT cn chng minh trthnh

( )

( )

( )

( )

( )

( )

2 2 2

2 2 22 2 2

1 1 18

2 1 2 1 2 1

a b c

a a b b c c

+ + ++ +

+ + +

. vi ( ), , 0;1a b c .

t ( ) ( )

( )

2 2

2 22

1 2 1

3 2 12 1

x x xf x

x xx x

+ + += =

++

, vi ( )0;1x .

Khi bt ng thc cn chng minh trthnh ( ) ( ) ( ) 8f a f b f c+ + Ta don ng thc xy ra khi 1

3a b c= = = . Phng trnh tip tuyn ca

thhm s ( )f x ti im1 16

;3 3

M

l 1 1 16 12 4'3 3 3 3

xy f x

+ = + =

.

Bng trc quan hnh hc ta thy th hm s ( )y f x= nm di tip

tuyn. trong khong (0 ; 1). Gi ta chng minh BT

( ) ( )12 4

, 0;13

xf x x

+

23 2

2

2 1 12 436 15 2 1 03 2 1 3

x x xx x xx x

+ + +

+ + ( ) ( )2

3 1 4 1 0, (0;1).x x x + Bt ng thc cui cng hin nhin l ng.

Suy ra ( ) ( ) ( ) ( )12 12

83

a b cf a f b f c

+ + ++ + = (pcm).

V d 3. (NHT BN 1997)Cho a, b, c l cc sthc dng. Chng minh rng

5

3

)(

)(

)(

)(

)(

)(22

2

22

2

22

2

++

++

++

++

++

+

cba

cba

bac

bac

acb

acb.

Li gii. BT cn chng minh c tnh thun nht, khng mt tng qut, gis

3a b c+ + = .Khi ( ), , 0;3a b c v bt ng thc cn chng minh trthnh( )

( )

( )

( )

( )

( )

2 2 2

2 2 22 2 2

3 2 3 2 3 2 3

53 3 3

a b c

a a b b c c

+ +

+ + +

2 2 2

1 1 1 3

2 6 9 2 6 9 2 6 9 5a a b b c c + +

+ + +

Hay 3( ) ( ) ( )5

f a f b f c+ + vi ( ) 21

2 6 9f x

x x=

+.


3/11

3

Phng trnh tip tuyn ti im 1x= ca thhm s ( ) 21

2 6 9f x

x x=

+

l ( )( ) ( ) ( )2 1 2 3

' 1 1 1 125 5 25

xy f x f x

+= + = + = .

Ta schng minh ( )21 2 3

, 0;32 6 9 25

xx

x x

+

+

( ) ( )3 3 22 1 3 0, 0;3x x x x + +

Theo BT AM GM th 3 3 21 3x x x+ + nn bt ng thc trn ng.

Do suy ra ( ) ( ) ( ) ( ) ( ) ( )2 3 2 3 2 3 3

25 5

a b cf a f b f c

+ + + + ++ + = .

V d4. Cho a, b, c l di ba cnh ca mt tam gic. Chng minh rng1 1 1 9 1 1 1

4a b c a b c a b b c c a

+ + + + +

+ + + + + .

Li gii.Khng mt tnh tng qut, gis 1a b c+ + = . V a, b, cl ba cnh ca mt

tam gic nn 1, , 0;2

a b c

.

Bt ng thc cn chng minh tng ng vi

( ) ( ) ( )4 1 4 1 4 1

9 91 1 1

f a f b f ca a b b c c

+ + + +

Vi ( ) 24 1 5 1 1

, 0;1 2

xf x x

x x x x

= =

.

Ta don ng thc xy ra khi 13

a b c= = = . V vy ta tm phng trnh

tip tuyn ca thhm s ( )f x ti im

1

;33M

l 18 3y x=

.Ta schng minh

( ) ( ) ( )2

2

5 1 1 118 3, 0; 3 1 2 1 0, 0;

2 2

xf x x x x x x

x x

=

.

Bt ng thc ny ng vi 10;2

x

.

Do ( ) ( ) ( ) ( )18 9 9f a f b f c a b c+ + + + = (pcm)

Du bng xy ra khi 13

a b c= = = , do du bng xy ra ca bt ng thc

ban u l a b c= =

.II. Sdng tnh thun nht

Mt bt ng thc (ng thc hay biu thc) c gi l c tnh thunnht i vi cc bin 1 2, ,..., na a a nu khi thay 1a bi 1ka , 2a bi 2ka , ..., na bi nka th bt ng thc (ng thc hay biu thc) khng thay i, vi k l sthcty , khc 0.


4/11

4

Mt bt ng thc (hay mt biu thc) c tnh thun nht i vi cc bin

1 2, ,..., na a a , khi gii c tht bin ph lm gim bin trong bt ng thc(hay mt biu thc) nhm n gin ha bi ton.

V d5.(thi i hc khi A nm 2009)

Chng minh rng vi mi sthc dng , ,x y z thomn ( ) 3x x y z yz+ + =

, ta c( ) ( ) ( ) ( ) ( ) ( )

3 3 33 5x y x z x y x z y z y z+ + + + + + + + .

Nhn xt. Ta c( ) 23 3x x y z yz x xy xz yz+ + = + + = ( ) ( ) ( ) ( )( )

2 2 2y z x y x z x y x z + = + + + + + .

Bt ng thc cn chng minh trthnh3 3

3 5.x y x z x y x z

y z y z y z y z

+ + + ++ +

+ + + +

t ,x y x za by z y z

+ += =

+ +,

Bi ton trthnh:Cho cc sthc dng , ,a b c thomn 2 2 1a b ab+ = . Chngminh rng 3 3 3 5a b ab+ + .

Li gii. Ta c2 2

2 2 2 2 2 21 2.2

a ba b ab a b a b

++ = + + 1+ +

Mt khc ( ) ( )2 2 24 2 4 2, 1ab a b a b a b ab + + + .

Khi ( )( )3 3 2 23 3 3a b ab a b a b ab ab a b ab+ + = + + + = + + 5 . (pcm)

V d6.(thi i hc khi A nm 2013)

Cho cc sthc dng , ,a b c tha mn iu kin ( )( ) 24a c b c c+ + = . Tm gi tr

nhnht ca biu thc3 3 2 2

3 3

32 32

( 3 ) ( 3 )

a b a bP

b c a c c

+= +

+ +.

Nhn xt. Ta c ( ) ( ) 24 1 1 4a b

a c b c cc c

+ + = + + =

.

Biu thc

3 3

2 2

3 3

32. 32.

3 3

a b

a bc cP

c cb a

c c

= + +

+ +

.

t ,a bx yc c

= = .

Bi ton trthnh: Cho cc sthc dng , ,x y z thomn 3xy x y+ + = . Tm gi

trnhnht ca biu thc( ) ( )

3 32 2

3 3

32 32

3 3

x yP x y

y x= + +

+ +

.

Li gii.Ta c


5/11

5

( ) ( )

3 32 2

3 3

32 1 1 32 1 1

2 2 2 23 3

x yP x y

y x

= + + + + + +

+ + 2 2 22

3 3

x yx y

y x

6 + +

+ +

2 24 3 8 9xy x y xy= + .

t t xy= , tgithit suy ra ( ]2 2 3 0 0;1t t t+ .

Xt hm s ( ) 24 3 8 9f t t t t= + trong khong (0 ; 1] c

( )2

2 2

4 4 3 8 9' 3

8 9 8 9

t t t t f t

t t t t

+= + =

+ +

( )

( ) ( ]

2

2 2

2 16 4470, 0;1

8 9 4 3 8 9

tt

t t t t t

+= <

+ + +

( ) ( )1 1 2f t f = . Du bng xy

ra khi v chkhi 1t= .

Vy gi trnhnht ca biu thc P l 1 2 , t khi a b c= = .

V d7. Cho cc sthc [ ], , 1;2a b c . Tm gi trln nht v gi trnhnht

ca2 2 2 2

22 2 2

4 2 153

4

a ac c b bc cP a a

a c bc c

+ + + = + +

+ .

Nhn xt.

2 2

22

4. 1 2. 153

41

a a b b

c c c cP a a

bacc

+ + +

= + +

+

.

t ,a bx yc c

= = ,z = athx, y 1 ;2

2

v [ ]1; 2z .

Bi ton trthnh: Cho cc sthc dng [ ]1

, ;2 , 1;22

x y z

. Chng minh

rng2 2

22

4 1 2 153

1 4

x x y yP z z

x y

+ + + = + +

+ .

Li gii.

Kho st ba hm s ( )2

2

4 1 1, ;2

1 2

x xf x x

x

+ + =

+ ; ( )

2 2 15 1, ;2

4 2

y yg y y

y

+ =

v

( ) [ ]2 3 , 1;2h z z z z= + . Suy ra37 3 3

Max 1; 1; .4 2 2P x y z a b c= = = = = = = 81

Min 2, 1 2, 110

P x y z a b c= = = = = = = .

V d8. Cho cc sthc dng , ,a b c . Chng minh rng3 3 3

3 3 3 3 3 31

a b c

a abc b b abc c c abc a+ +

+ + + + + +.


6/11

6

Nhn xt. Bin i bt ng thc cn chng minh2 2 2

2 2 2 2 2 2 11 1 1

a b c

bc ca ab

a b b c c a

bc ca ca ab ab bc

+ +

+ + + + + +

.

t , ,a b cx y zb c a

= = = ,

Bi ton trthnh:Cho cc sthc dng , ,x y z . Chng minh rng2 2 2

2 2 21

x y z

x xz yz y yz zx z zy xy+ +

+ + + + + +.

p dng BT

( ) ( ) ( )

22 2 2 2 2 22 a b ca b c a b c

x y z a b cx y z x y z x y z

+ + + + + + + + + +

+ + , ng thc xy ra khi

.a b c

x y z= =

Ta c

( )22 2 2

2 2 2 2 2 21

x y zx y z

x xz yz y yz zx z zy xy x xz yz y yz zx z zy xy

+ ++ + =

+ + + + + + + + + + + + + +.

iu phi chng minh.

Ch .Khi gp cc biu thc c dng ( , )( , )

f x yP

g x y= , trong ( , )f x y , ( , )g x y l cc

biu thc ng cp th ta c tht ( 0)x ty y= hayx

t

y

= a Pvhm mt

bin t.

V d9. Cho x, y tha mn , 0; 1x y xy y> .Tm GTLN ca biu thc

2 2

2

6( )3

x y x yP

x yx xy y

+ =

+ +

(thi H khi D nm 2013)

Li gii.Do2

2 2

1 1 1 1 1 1 1, 0; 1 0

4 2 4

x yx y xy y

y y y y y

> < = =

.

t xty

= , suy ra 104

t< . Ta c2

1 2

6( 1)3

t tP

tt t

+ =

+ +

.

Xt hm s2

1 2 1( ) (0 )

6( 1) 43

t tf t t

tt t

+ = <

+ +

;22 3

7 3 1'( )

2( 1)2 ( 3)

tf t

tt t

=

+ +

Vi 104

t< th

222 3

7 3 7 3 1 1 13 ( 1) 3 3; 1 1 ;

2( 1) 26 3 32 ( 3)

t tt t t t t

tt t

+ = + < + > > > >

+ +

.


7/11

7

Suy ra 1 1'( ) 023

f t > > , tc l hm s ( )f t ng bin trn1

0;4

.

Do 1 5 7( )4 3 30

P f t f

= = +

. Khi 1 5 7, 2 th2 3 30

x y P= = = + .

Vy GTLN c

a Pl

5 7

3 30+

.III. Phng php tham sho

Khi gp cc hm snhiu bin ta i kho st hm stheo mt bin, cc bin cnli xem nhl tham s.Vic chng minh bt ng thc vi bin s trong mt on no , ta quy vchng minh mt bt ng thc n gin hn ng vi bin snhn ti mt vigi trcth(thng l cc im nt ca on ).

Nhn xt 1. Cho ( )f x mx n= + . Khi ta c1)

min { } { }( ), ( ) ( ) max ( ), ( )f a f b f x f a f b vi mi [ ]; .x a b 2) Nu ( ) 0; ( ) 0f a f b th ( ) 0f x vi mi [ ]; .x a b 3)

Nu ( ) 0; ( ) 0f a f b th ( ) 0f x vi mi [ ]; .x a b Nhn xt 2. Cho ( )2( ) 0f x mx nx p m= + + . Khi ( )f x nhn gi trln nht,

gi trnhnht tix = ahocx = bhocx =2

n

m .

Nhn xt 3.1) Nu ( )f x l hm li trn [a ; b] (tc lf(x) < 0 trn [a ; b]) th

{ }( ) min ( ); ( )f x f a f b vi mi [ ]; .x a b .2) Nu ( )f x l hm lm trn [a ; b] (tc lf(x) > 0 trn [a ; b]) th

{ }( ) max ( ); ( )f x f a f b vi mi [ ]; .x a b

V d10. Cho x, y, z, t thuc[0 ; 1]. Chng minh rng(1 x)(1 y) (1 z)(1 t) +x+y+z+ t 1.

Li gii.Bin i BT cn chng minh thnh(1 x)(1 y) (1 z)(1 t) +x+y+z+ t1 0.

Coi vtri l a thc dngf (x) = mx + n.Theo Nhn xt 1 th( ) ( ){ }( ) min 0 ; 1f x f f vi mi [ ]0;1.x

Ta c (1) 0 ; (0) (1 )(1 )(1 ) 1.f y z t f y z t y z t= + + = + + +

Xt hm g(y) : =f(0) th g(y) min {g(0) ; g(1)} vi miy [0 ; 1].Ta c g(1) =z+ t0 ; g(0) = ( 1 y)(1 z)(1 t) +z + t1 =zt 0. Do g(y) 0 vi miy [0 ; 1]. T suy rafx) 0 (pcm).ng thc xy ra chng hn tix= 1 ;y = z= 0.

V d11. Cho ba sdng x, y, z thomn iu kin x + y + z = 1.Chng minh rng

xy + yz + zx2xyz 727

.


8/11

8

Li gii. BT cn chng minh tng ng vi

x(y + z) + yz 2xyz 7 0.27

Ty + z = 1 x suy ra

yz(1 2x) + x(1 x) 7 0.

27

Nhn thy ( ) ( )2 2

10

4 4

y z xyz

+ = . t yz = t, xt hm s

7( ) (1 2 ) (1 )

27f t t x x x= + trn on ( )

21

0;4

x

. Theo nhn xt 1 th

( )2

1( ) max (0);

4

xf t f f

m 7(0) (1 ) 027

f x x= vi mixthuc [0 ; 1]. T suy ra pcm.

ng thc xy ra khix = y = z= 1.3

V d12. Cho cc sdng x, y, z thomn iu kinx + y + z= 1.Chng minh rng 3 3 34( ) 15 1.x y z xyz+ + + Li gii. Ta c

( ) ( )( )

( ) ( )( )( ) ( )

33 3 3 3

33

33

4( ) 15 1 4 3 4 15 1

4 1 3 1 4 15 1

27 12 4 4 1 1.

.

x y z xyz x y xy x y z xyz

z xy z z xyz

xy z z z

+ + + = + + + +

= + +

= + +

Tgithit suy ra2 2 21

0 .2 4

x y zxy

+ t xy = t, xt

vi ( )2

10;

4

zt

. Ta c

f(0) = 3(2z1)20 ;

( ) ( )( ) ( )

2 222

1 1 327 12 3 12 12 3 1 0.

4 4 4

z z zf z z z z

= + + =

Vyf(t) 0 vi

( )2

10;

4

zt

. Suy ra BT cn chng minh. ng thc xy ra khi v chkhi x =

y = z= 1.3

V d13.Cho cc sa, b, c khng m. Chng minh rng

( ) ( ) ( )2 2 2

3 max{ a ; b ; c3

a b cabc b c a

+ + .

( ) ( )33( ) 27 12 4 4 1 1.f t t z z z= + +


9/11

9

Li gii. Do vai tr a, b, cnhnhau, khng gim tng qut gis a b c. Tacn chng minh

( )2

3 a3

a b cabc c

+ +

t ( )2

3( ) a

3

a x cf x acx c

+ += . Khi

2 2

3 5 5 5

2''( ) . 0.

9

a cf x

x a c

= > nn

f(x) lm trn [a ; c]. Theo tnh cht 3 thf(x) max {f(a) ;f(c)}.

( )3 3 3 32 2 2 22 1

( ) 2 23 3 3

2 2 0.

a cf a a c ac a c c a c a c a c ac

ac ac

= + = + + + + + +

+ =

V vyf(a) 0. Tng tf(c) 0. Do f(x) 0 vi mixthuc [a ; c].

Suy ra pcm,V d14. Cho cc sa, b, c thuc[0 ; 1]. Chng minh rng

(1 )(1 )(1 ) 1.1 1 1

a b ca b c

b c c a a b+ + +

+ + + + + +

Li gii. Gisa= max {a, b, c}. Khi

VT (1 )(1 )(1 ) 11

a b ca b c

b c

+ ++

+ +vi mi thuc [0 ; 1].

t ( ) (1 )(1 )(1 ) 11

x b cf x x b c

b c

+ += +

+ + th f(1) = 0

2 2

2 2

( ) ( )(0) (1 )(1 ) 1

1 1( ) 1

2 0.1

b c bc b c b c bcf b c

b c b cb c

b c bc

b c

+ + + = + =

+ + + +

+ +

0, 1 z > 0 nn g(0) 3

1 1 8.

3 27

y z z y+ + + =

g(1) = (1 z) + (z21)+ y + y2z y2yz2 = (yz)(1 z)(1 y) 3

2 2 8

3 27.z

V vy g(t) 827

, nnf(x) 827

(pcm).

BI TP

1.Cho bn sthc khng m , , ,a b c d tha mn iu kin 4a b c d + + + = . Chngminh rng

2 2 2 2

1

5 3 5 3 5 3 5 3 2

a b c d

a b c d + + +

+ + + +.

2.Cho , ,a b c l cc sdng v 3a b c+ + = . Chng minh rng

( ) ( ) ( )

2 2 2

2 2 22 2 2

9 9 95

2 2 2

a b c

a b c b c a c a b

+ + ++ +

+ + + + + +. (Trung Quc- 2006)

3. Cho 3, ,4

a b c v 1a b c+ + = . Chng minh rng 2 2 29

1 1 1 10

a b c

a b c+ +

+ + +.

4. Cho , , , 0a b c d > v 4a b c d + + + = . Chng minh rng :3 3 3 3

4

2 2 2 2 27

a b c d

a b c d

+ + +

+ + + + .

5. Cho a, b, c l cc skhng m tha mn 0a b c+ + > . Chng minh rng

( ) ( ) ( )

2 2 2

2 2 22 2 2

1

35 5 5

a b c

a b c b c a c a b+ +

+ + + + + +

.

6. Cho a, b, c l cc skhng m tha mn 0a b c+ + > . Chng minh rng

( ) ( ) ( )

2 2 2

2 2 22 2 21 22 32 2 2

a b c

a b c b c a c a b + +

+ + + + + +.

7.Cho , ,x y z l ba sthc thuc on [ ]1;4 v ,x y x z . Tm gi trnhnht

ca biu thc2 3

x y zP

x y y z z x= + +

+ + +. (thi i hc khi A nm 2011).

8. : Cho x, y tha mn: 2 2 1x y+ = . Tm GTLN, GTNN ca biu thc:


11/11

11

2

2

6

1 2 2

x xyP

xy y

+=

+ +(thi H khi B -2008)

9.Cho 3 sx, y, z tha mn 3 3 3 3 1x y z xyz+ + = . Tm GTNN ca biu thc.2 2 2

P x y z= + + (thi chn i tuyn dthi IMO ca Innxia -2009)

10.Cho cc sthc dng a, b tha mn: 2 22( ) ( )( 2)a b ab a b ab+ + = + + . Tm

GTNN ca biu thc3 3 2 2

3 3 2 24 9a b a b

Pb a b a

= + +

(thi H khi B -2011).

11. Cho a, b, cl cc sdng thomn iu kin a + b + c= 1. Chng minhrng

7(ab + bc + ca) 2 + 9abc.

12. Cho a, b, c, d , ethuc [p ; q] vi q> p > 0. Chng minh rng2

1 1 1 1 1( ) 25 6

p qa b c d e

a b c d e q p

+ + + + + + + + +

13.Cho cc sx, y, zdng v thomn diu kinx + y + z= 1. Chng minhrng

a) 9xyz+ 1 4(xy + yz + zx)b)

5(x2+y2+z2) 6(x3+y3+z3) + 1.

14.Cho nsthuc [0 ; 1] vi n2. Chng minh rng1 2

1 2

1 2

... (1 )(1 )...(1 ) 11 1 1

n

n

n

a a a a a aS a S a S a

+ + + + + + +

vi 1 2 ... .nS a a a= + + +

phuong phap chung minh bdt 1

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