19 PP Chung Minh Bat Dang Thuc

8/6/2019 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:

1


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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

2


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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:

3


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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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=(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:

11


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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

12


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( )( )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

13


14/37


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 ++


18/37

x + y = 1

MN

O

MKH

M

x

y


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

18


19/37


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

19


20/37



21/37


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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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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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

23


24/37


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

24


25/37


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


26/37


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 :

26


27/37


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:

27


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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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* 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

29


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=(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

30


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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


32/37


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

33


34/37


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

34


35/37


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

35


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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

36


37/37


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

19 PP Chung Minh Bat Dang Thuc

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Transcript of 19 PP Chung Minh Bat Dang Thuc