Chuyen de Bat Dang Thuc Luong Giac (Chuong 1)

8/14/2019 Chuyen de Bat Dang Thuc Luong Giac (Chuong 1)

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Trng THPT chuyn L T Trng Cn Th Bt ng thc lng gicChng 1Cc bc u cs

The Inequalities Trigonometry 3

Chng 1 :

CC BC U CS

bt u mt cuc hnh trnh, ta khng th khng chun b hnh trang ln ng.Ton hc cng vy. Mun khm ph c ci hay v ci p ca bt ng thc lnggic, ta cn c nhng vt dng chc chn v hu dng, chnh l chng 1: Ccbc u cs.

Chng ny tng qut nhng kin thc cbn cn c chng minh bt ng thclng gic. Theo kinh nghim c nhn ca mnh, tc gi cho rng nhng kin thc ny ly cho mt cuc hnh trnh.

Trc ht l cc bt ng thc i s cbn ( AM GM, BCS, Jensen, Chebyshev) Tip theo l cc ng thc, bt ng thc lin quan cbn trong tam gic. Cui cngl mt snh l khc l cng cc lc trong vic chng minh bt ng thc (nh lLargare, nh l v du ca tam thc bc hai, nh l v hm tuyn tnh )

Mc lc :1.1. Cc bt ng thc i s cbn 4

1.1.1. Bt ng thc AM GM............................................... 4

1.1.2. Bt ng thc BCS.. 81.1.3. Bt ng thc Jensen.... 131.1.4. Bt ng thc Chebyshev..... 16

1.2. Cc ng thc, bt ng thc trong tam gic.. 191.2.1. ng thc... 191.2.2. Bt ng thc..... 21

1.3. Mt s nh l khc. 221.3.1. nh l Largare ... 221.3.2. nh l v du ca tam thc bc hai.. 251.3.3. nh l v hm tuyn tnh.. 28

1.4. Bi tp.. 29


2/28



1.1. Cc bt ng thc i s cbn :

1.1.1. Bt ng thc AM GM :

Vi mi s thc khng m naaa ,...,, 21 ta lun c

nn

n aaan

aaa...

...21

21

+++

Btng thcAM GM(Arithmetic Means Geometric Means) l mt btng thcquen thuc v c ng dng rt rng ri.y l btng thc m bn c cn ghi nh rrng nht, n s l cng c hon ho cho vic chng minh cc btng thc. Sau y lhai cch chng minh btng thc ny m theo kin ch quan ca mnh, tcgi chorng l ngngn v hay nht.

Chng minh :Cch 1 : Quy np kiu Cauchy

Vi 1=n bt ng thc hin nhin ng. Khi 2=n bt ng thc tr thnh

( ) 02

2

212121

+

aaaaaa

(ng!)

Gi s bt ng thc ng n kn = tc l :

kk

k aaak

aaa...

...21

21

+++

Ta s chng minh n ng vi kn 2= . Tht vy ta c :

( ) ( ) ( )( )

( )( )

kkkk

kkkk

kk

kkkkkkkk

aaaaa

k

aaakaaak

k

aaaaaa

k

aaaaaa

22121

22121

2212122121

......

......

......

2

......

+

++

++++

=

++++++

+++++++

Tip theo ta s chng minh vi 1= kn . Khi :

( ) 1 121121

1121

1121121

1121121

...1...

...

............

=

+++

=

++++

kkk

kk

k kkk

kkk

aaakaaa

aaak

aaaaaakaaaaaa

Nh vy bt ng thc c chng minh hon ton.ng thc xy ra naaa === ...21

Cch 2 : ( ligii ca Polya )


3/28



Gin

aaaA

n+++=

...21

Khi bt ng thc cn chng minh tng ng vin

n Aaaa ...21 (*)

R rng nu Aaaa n====

...21 th (*) c du ng thc. Gi s chng khng bngnhau. Nh vyphi c t nht mt s, gi s l Aa 2

tc l 21 aAa =+= AaAaaaAaaAaaaa

2121 '' aaaa >

nn aaaaaaaa ...''... 321321


4/28



Li gii :

Ta lun c : ( ) CBA cotcot =+

1cotcotcotcotcotcot

cotcotcot

1cotcot

=++

=+

ACCBBA

CBA

BA

Khi :( ) ( ) ( )

( ) ( )

3cotcotcot

3cotcotcotcotcotcot3cotcotcot

0cotcotcotcotcotcot

2

222

++

=++++

++

CBA

ACCBBACBA

ACCBBA

Du bng xy ra khi v ch khi ABC u.

V d 1.1.1.3.

CMR vi mi ABCnhn v *Nn ta lun c :

2

1

3tantantan

tantantan

++

++nnnn

CBA

CBA

Li gii :

Theo AM GM ta c :

( ) ( )

( ) ( ) 21

33

3 3

33

3333tantantan3tantantan

tantantan

tantantan3tantantan3tantantan

=++

++

++

++=++

nnn

nnn

nnnnn

CBACBA

CBA

CBACBACBA

pcm.

V d 1.1.1.4.

Cho a,b l hai sthc tha :0coscoscoscos ++ baba

CMR : 0coscos + ba

Li gii :

Ta c :

( )( ) 1cos1cos1

0coscoscoscos

++

++

ba

baba

Theo AM GM th :


5/28



( ) ( )( )( )

0coscos

1cos1cos12

cos1cos1

+

+++++

ba

baba

V d 1.1.1.5.

Chng minh rng vi mi ABC nhn ta c :

2

3

2sin

2sin

2sin

2sin

2sin

2sin

3

2

2cos

2cos

coscos

2cos

2cos

coscos

2cos

2cos

coscos+

++++

ACCBBA

AC

AC

CB

CB

BA

BA

Li gii :

Ta c

=

=

BABA

BA

BA

AA

A

A

cotcot4

3

2sin

2sin

2cos

2cos4

coscos4

3

2cot2sin

2cos2

cos

Theo AM GM th :

+

+

BABA

BA

BA

BABA

BA

BA

cotcot4

3

2sin

2sin

3

2

2cos

2cos

coscos

2

cotcot4

3

2sin

2sin

2cos2cos4

coscos4

32

Tng t ta c :

+

+

AC

AC

AC

AC

CBCB

CB

CB

cotcot4

3

2sin2sin3

2

2cos

2cos

coscos

cotcot4

3

2sin

2sin

3

2

2cos

2cos

coscos

Cng v theo v cc bt ng thc trn ta c :


6/28



( )ACCBBAACCBBA

AC

AC

CB

CB

BA

BA

cotcotcotcotcotcot

2

3

2

sin

2

sin

2

sin

2

sin

2

sin

2

sin

3

2

2cos

2cos

coscos

2cos

2cos

coscos

2cos

2cos

coscos

+++

++

++

2

3

2sin

2sin

2sin

2sin

2sin

2sin

3

2+

++=

ACCBBApcm.

Bc u ta mi ch c btng thcAM GMcng cc ng thc lnggic nnsc nh hng n cc btng thc cn hn ch. Khi ta kt hpAM GMcngBCS,Jensen hay Chebyshev th n thc s l mtv kh ng gm cho cc btng thclnggic.

1.1.2. Bt ng thc BCS :

Vi hai b s ( )naaa ,...,, 21 v ( )nbbb ,...,, 21 ta lun c :

( ) ( )( )2222

1

22

2

2

1

2

2211 ......... nnnn bbbaaabababa +++++++++

Nu nhAM GMl cnh chim u n trong vic chng minh btng thc thBCS (Bouniakovski Cauchy Schwartz) li l cnh tay phi ht sc c lc. Vi

AM GMta lun phi ch iu kin cc bin l khng m, nhng i vi BCS ccbin khng b rng buc bi iu kin , chcn l sthc cng ng. Chng minh btng thc ny cng rtngin.

Chng minh :

Cch 1 :

Xt tam thc :

( ) ( ) ( )22222

11 ...)( nn bxabxabxaxf +++=

Sau khi khai trin ta c :

( ) ( ) ( )22

2

2

12211222

2

2

1 ......2...)( nnnn bbbxbababaxaaaxf ++++++++++= Mt khc v Rxxf 0)( nn :

( ) ( )( ) +++++++++ 2222

1

22

2

2

1

2

2211 .........0 nnnnf bbbaaabababa pcm.

ng thc xy ran

n

b

a

b

a

b

a=== ...

2

2

1

1 (quy c nu 0=ib th 0=ia )

Cch 2 :


7/28



S dng bt ng thc AM GM ta c :

( )( )2222

1

22

2

2

1

22

2

2

1

2

22

2

2

1

2

......

2

......nn

ii

n

i

n

i

bbbaaa

ba

bbb

b

aaa

a

++++++

+++

++++

Cho i chy t 1 n n ri cng v c n bt ng thc li ta c pcm.y cng l cch chng minh ht sc ngngn m bn c nn ghi nh!

By givi stip sc caBCS,AM GMnhc tip thm ngun sc mnh, nhh mc thm cnh, nhrng mc thm vy,pht huy hiu qu tm nh hng ca mnh.Hai btng thc ny b p bsung htrcho nhau trong vic chng minh btngthc. Chng lng long nht th, song kim hp bch cngph thnh cng nhiubi ton kh.

Trm nghe khng bng mt thy, ta hyxtcc v d thy r iu ny.

V d 1.1.2.1.

CMR vi mi ,,ba ta c :

( )( )2

21cossincossin

++++

baba

Li gii :

Ta c :

( )( ) ( )( )

( ) ( )( ) ( )12cos12sin12

1

2

2cos12sin

22

2cos1

coscossinsincossincossin 22

++++=

++

++

=

+++=++

abbaab

abba

abbaba

Theo BCS ta c :

( )2cossin 22 BAxBxA ++

p dng ( )2 ta c :

( ) ( ) ( ) ( ) ( )( ) ( )31112cos12sin 2222 ++=++++ baabbaabba Thay ( )3 vo ( )1 ta c :

( )( ) ( )( )( ) ( )41112

1cossincossin 22 ++++++ baabba

Ta s chng minh bt ng thc sau y vi mi a, b :

( )( )( ) ( )52

11112

12

22

++++++

babaab


8/28



Tht vy :

( ) ( )( )

( )( )2

211

24111

2

1

22

15

2222

2222

++++

++

+++++

baba

abbaba

ab

( )( ) ( ) ( ) ( )62

1111

2222 +++

++ba

ba

Theo AM GM th ( )6 hin nhin ng ( )5 ng.T ( )1 v ( )5 suy ra vi mi ,,ba ta c :

( )( )2

21cossincossin

++++

baba

ng thc xy ra khi xy ra ng thi du bng ( )1 v ( )6

( )

++=

=

+=

=

=+

=

Zkkab

baarctg

ba

abbatg

ba

abba

ba

2121

12cos1

2sin

22

V d 1.1.2.2.

Cho 0,, >cba v cybxa =+ cossin . CMR :

33

222 11sincos

ba

c

bab

y

a

x

+++

Li gii :Bt ng thc cn chng minh tng ng vi :

( )*cossin

11cos1sin1

33

222

33

222

ba

c

b

y

a

x

ba

c

bab

y

a

x

++

++

+

Theo BCS th :

( ) ( )( )222

1

2

2

2

1

2

2211 bbaababa +++

vi

==

==

bbbaab

bya

axa

21

21

;

cos;sin

( ) ( )23322

cossincossin

ybxabab

y

a

x++

+

do 033 >+ ba v ( )*cossin =+ cybxa ng pcm.


9/28



ha

x

yz

N

Q

P

A

B C

M

ng thc xy ra22

2

2

1

1 cossin

b

y

a

x

b

a

b

a==

+=

+=

=+

=

33

2

33

2

22

cos

sin

cossin

cossin

ba

cby

ba

cax

cybxa

b

y

a

x

V d 1.1.2.3.

CMR vi mi ABC ta c :

R

cbazyx

2

222++

++

vi zyx ,, l khong cch t im M bt k nm bn trong ABC n ba cnhABCABC ,, .

Li gii :

Ta c :

( )

++++=++

=++

=++

++=

cba

cbacba

abc

ABC

MCA

ABC

MBC

ABC

MAB

MCAMBCMABABC

h

z

h

y

h

xhhhhhh

h

x

h

y

h

z

S

S

S

S

S

S

SSSS

1

1

Theo BCSth :

( )cba

cba

cba

c

c

b

b

a

a hhhh

z

h

y

h

xhhh

h

zh

h

yh

h

xhzyx ++=

++++++=++

m BahAchCbhCabahS cbaa sin,sin,sinsin21

21 =====

( )R

ca

R

bc

R

abAcCbBahhh

cba222

sinsinsin ++=++=++

T suy ra :

++

++

++R

cba

R

cabcabzyx

22

222

pcm.


10/28



ng thc xy ra khi v ch khi ABCzyx

cba

==

==u v M l tm ni tip ABC .

V d 1.1.2.4.

Chng minh rng :

+

2;08sincos 4

xxx

Li gii :

p dng bt ng thc BCS lin tip 2 ln ta c :

( ) ( )( )( )( ) ( )( )

4

2222222

2224

8sincos

8sincos1111

sincos11sincos

+

=+++

+++

xx

xx

xxxx

ng thc xy ra khi v ch khi4

=x .

V d 1.1.2.5.

Chng minh rng vi mi sthc a vx ta c

( ) 11

cos2sin12

2

+

+

xaxax

Li gii :

Theo BCS ta c :

( )( ) ( ) ( ) ( )

( )( ) ( )

( ) 11

cos2sin1

1cos2sin1

21421

cossin21cos2sin1

2

2

2222

42242

2222222

+

+

++

++=++=

+++

xaxaa

xaxax

xxxxx

aaxxaxax

pcm.


11/28



1.1.3. Bt ng thc Jensen :

Hm s )(xfy = lin tc trn on [ ]ba, v n im nxxx ,...,, 21 ty trn on

[ ]ba, ta c :

i) 0)('' >xf trong khong ( )ba, th :

++++++

n

xxxnfxfxfxf nn

...)(...)()( 2121

ii) 0)(''


12/28



V d 1.1.3.1.

Chng minh rng vi mi ABC ta c :

2

33sinsinsin ++ CBA

Li gii :

Xt xxf sin)( = vi ( );0x

Ta c ( );00sin)('' =

2;00

cos

sin2''3

xx

xxf . T theo Jensen th :

==

++

+

+

3

6sin3

3

2223222

CBA

fC

fB

fA

f pcm.

ng thc xy ra khi v ch khi ABC u.

V d 1.1.3.3.


21

222222

32

tan2

tan2

tan

+

+

CBA

Li gii :


13/28



Xt ( ) ( ) 22tanxxf = vi

2;0

x

Ta c ( ) ( )( ) ( ) ( ) 1221221222 tantan22tantan122' + +=+= xxxxxf

( ) ( )( )( ) ( )( )( ) 0tantan1122tantan112222'' 2222222 >++++= xxxxxf Theo Jensen ta c :

=

=

++

+

+

2122

36

33

2223222

tg

CBA

fC

fB

fA

f pcm.


V d 1.1.3.4.

Chng minh rng vi mi ABC ta c :3

2

3

2tan

2tan

2tan

2sin

2sin

2sin ++++++

CBACBA

Li gii :

Xt ( ) xxxf tansin += vi

2;0

x

Ta c ( )( )

>

=

2;00

cos

cos1sin''

4

4 x

x

xxxf

Khi theo Jensen th :

+=

+=

++

+

+

3

2

3

6tan

6sin3

3

2223222

CBA

fC

fB

fA

f pcm.


V d 1.1.3.5.

Chng minh rng vi mi ABC nhn ta c :

( ) ( ) ( )2

33

sinsinsin

3

2sinsinsin

CBACBA

Li gii :

Ta c


14/28



++++

+=++

CBACBA

CBACBA

222

222

sinsinsinsinsinsin

coscoscos22sinsinsin

v2

33sinsinsin ++ CBA

2

33sinsinsin2 ++


15/28



Chng minh :

Bng phn tch trc tip, ta c ng thc :

( ) ( )( ) ( )( ) 0.........1,

21212211 =+++++++++ =

n

ji

jijinnnnbbaabbbaaabababan

V hai dy naaa ,...,, 21 v nbbb ,...,, 21 n iu cng chiu nn ) ) 0 jiji bbaa

Nu 2 dy naaa ,...,, 21 v nbbb ,...,, 21 n iu ngc chiu th bt ng thc i

chiu.

V d 1.1.4.1.


3

++

++

cba

cCbBaA

Li gii :

Khng mt tnh tng qut gi s :CBAcba

Theo Chebyshev th :

33

333

=++++++

++

++

++

CBAcbacCbBaA

cCbBaACBAcba


V d 1.1.4.2.

Cho ABC khng c gc t vA, B, Co bng radian. CMR :

( ) ( )

++++++

C

C

B

B

A

ACBACBA

sinsinsinsinsinsin3

Li gii :

Xt ( )x

xxf

sin= vi

2;0

x

Ta c ( )( )

=

2;00

tancos'

2

x

x

xxxxf


16/28



Vy ( )xf nghch bin trn

2;0

Khng mt tng qut gi s :

C

C

B

B

A

ACBA

sinsinsin

p dng bt ng thc Chebyshev ta c :

( ) ( )++

++++ CBA

C

C

B

B

A

ACBA sinsinsin3

sinsinsinpcm.


V d 1.1.4.3.


3tantantan

coscoscossinsinsin CBA

CBACBA

++++

Li gii :

Khng mt tng qut gi s CBA

CBA

CBA

coscoscos

tantantan

p dng Chebyshev ta c :

3

tantantan

coscoscos

sinsinsin3

costancostancostan

3

coscoscos

3

tantantan

CBA

CBA

CBA

CCBBAACBACBA

++

++

++

++

++

++

M ta li c CBACBA tantantantantantan =++ pcm.


V d 1.1.4.4.


( )CBA

CBACBA

coscoscos

2sin2sin2sin

2

3sinsinsin2

++

++++

Li gii :

Khng mt tng qut gi s cba


17/28



CBA

CBA

coscoscos

sinsinsin

Khi theo Chebyshev th :

( )CBA

CBACBA

CCBBAACBACBA

coscoscos

2sin2sin2sin

2

3sinsinsin2

3

cossincossincossin

3

coscoscos

3

sinsinsin

++

++++

++

++

++

pcm.ng thc xy ra khi v ch khi ABC u.

1.2. Cc ng thc bt ng thc trong tam gic :

Sau y l hu ht nhng ng thc, btng thc quen thuc trong tamgic v tronglnggic c dng trong chuyn ny hoc rt cn thit cho qu trnh hc ton cabn c. Cc bn c th dng phn ny nhmt t in nh tra cu khi cn thit.Haybn c cng c thchng minh ttc cc ktqu nhl bi tp rn luyn.Ngoi ra ticng xin nhc vi bn c rng nhng kin thc trong phn ny khi p dng vo bi tpu cn thitc chng minh li.

1.2.1. ng thc :

RC

c

B

b

A

a

2sinsinsin ===

Cabbac

Bcaacb

Abccba

cos2

cos2

cos2

222

222

222

+=

+=

+=

AbBac

CaAcb

BcCba

coscos

coscos

coscos

+=

+=

+=

( ) ( ) ( )

( )( )( )cpbpapp

rcprbprap

prCBARR

abc

CabBcaAbc

hchbhaS

cba

cba

=

===

===

===

===

sinsinsin24

sin

2

1sin

2

1sin

2

1

.2

1.

2

1.

2

1

2


18/28



4

22

4

22

4

22

2222

2222

2222

cbam

bac

m

acbm

c

b

a

+=

+=

+=

ba

Cab

l

ac

Bca

l

cb

Abc

l

c

b

a

+=

+=

+=

2cos2

2cos2

2cos2

( )

( )

( )

2sin

2sin

2sin4

2tan

2tan

2tan

CBAR

Ccp

Bbp

Aapr

=

=

=

=

+

=+

+

=+

+

=+

2tan

2tan

2tan

2tan

2tan

2tan

AC

AC

ac

ac

CB

CB

cb

cb

BA

BA

ba

ba

S

cbaCBA

S

cbaC

SbacB

S

acbA

4cotcotcot

4cot

4cot

4cot

222

222

222

222

++=++

+=

+=

+=

( )( )

( )( )

( )( )ab

bpapC

ca

apcpB

bc

cpbpA

=

=

=

2sin

2sin

2sin

( )

( )

( )ab

cppC

ca

bppB

bc

appA

=

=

=

2cos

2cos

2cos

( )( )( )

( )( )

( )

( )( )

( )cppbpapC

bpp

apcpB

appcpbpA

=

=

=

2tan

2tan

2tan

( )

CBACBA

R

rCBACBA

CBACBA

CBACBA

R

pCBACBA

coscoscos21coscoscos

12

sin2

sin2

sin41coscoscos


sinsinsin42sin2sin2sin

2cos

2cos

2cos4sinsinsin

222

222

=++

+=+=++

+=++

=++

==++


19/28



1cotcotcotcotcotcot

12

tan2

tan2

tan2

tan2

tan2

tan

2cot

2cot

2cot

2cot

2cot

2cot

tantantantantantan

=++

=++

=++

=++

ACCBBA

ACCBBA

CBACBA

CBACBA

( ) ( ) ( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )

( ) kCkBkAkCkBkA

kCkBkAkCkBkA

Ck

Bk

Ak

Ck

Bk

Ak

Ak

Ck

Ck

Bk

Bk

Ak

kAkCkCkBkBkA

kCkBkAkCkBkA

kCkBkAkCkBkA

Ck

Bk

AkCkBkAk

kCkBkAkCkBkA

Ck

Bk

AkCkBkAk

k

k

k

k

k

k


coscoscos211coscoscos

212cot

212cot

212cot

212cot

212cot

212cot

12

12tan2

12tan2

12tan2

12tan2

12tan2

12tan

1cotcotcotcotcotcot

tantantantantantan

coscoscos4112cos2cos2cos

212sin

212sin

212sin41112cos12cos12cos

sinsinsin412sin2sin2sin

212cos

212cos

212cos4112sin12sin12sin

1222

222

1

+

+

+=++

+=++

+++=+++++

=++++++++

=++

=++

+=++

++++=+++++

=++

+++=+++++

1.2.2. Bt ng thc :

acbac

cbacb

bacba

+


20/28



1cotcotcot

9tantantan

4

9sinsinsin

4

3coscoscos

222

222

222

222

++

++

++

++

CBA

CBA

CBA

CBA

2cot

2cot

2cot

12tan2tan2tan

2sin

2sin

2sin

2cos

2cos

2cos

222

222

222

222

CBA

CBA

CBA

CBA

++

++

++

++

33

1cotcotcot

33tantantan

8

33sinsinsin

8

1coscoscos

CBA

CBA

CBA

CBA

332

cot2

cot2

cot

33

1

2

tan

2

tan

2

tan

8

1

2sin

2sin

2sin

8

33

2cos

2cos

2cos

AAA

AAA

CBA

CBA

1.3. Mt s nh l khc :

1.3.1. nh l Lagrange :

Nu hm s ( )xfy = lin tc trn on [ ]ba ; v c o hm trn khong ( )ba ; th tn ti 1 im ( )bac ; sao cho :

( ) ( ) ( )( )abcfafbf = '

Ni chung vi kin thc THPT, ta ch c cng nhn nh l ny m khng chng minh.Vchng minh ca n cn n mt skin thc ca ton cao cp. Ta chcn hiu cchdng n cng nhng iu kin i km trong cc trng hp chng minh.

V d 1.3.1.1.

Chng minh rng baRba


21/28



Xt ( ) ( ) xxfxxf cos'sin == Khi theo nh l Lagrange ta c

( ) ( ) ( ) ( )

abcabab

cabafbfbac

=

cossinsin

cos:;:

pcm.

V d 1.3.1.2.

Vi ba


22/28



CMR nu 0>x thxx

xx

+>

++

+

11

1

11

1

Li gii :

Xt ( ) ( )( ) 0ln1ln1

1ln >+=

+= xxxx

xxxf

Ta c ( ) ( )1

1ln1ln'

++=

xxxxf

Xt ( ) ttg ln= lin tc trn [ ]1; +xx kh vi trn ( )1; +xx nn theo Lagrange th :

( )( )( )

( )

( ) ( ) 01

1ln1ln'

1

1'

1

ln1ln:1;

>+

+=

+>=

+

++

xxxxf

xcg

xx

xxxxc

vi > 0x ( )xf tng trn ( )+;0

( ) ( )

xx

xx

xx

xxxfxf

+>

++

+>

++>+

+

+

11

1

11

11ln

1

11ln1

1

1

pcm.

V d 1.3.1.5.

Chng minh rng + Zn ta c :

1

1

1

1arctan

22

1222+

++

++ nnnnn

Li gii :

Xt ( ) xxf arctan= lin tc trn [ ]1; +nn

( )21

1'

xxf

+= trn ( ) ++ Znnn 1;

Theo nh l Lagrange ta c :( ) ( )

( ) ( )( )

( )( )

++=

+

++

+=+=

+

+

+=+

1

1arctan

1

1

11

1arctanarctan1arctan

1

1

1

1':1;

22

2

nnc

nn

nnnn

c

nn

nfnfcfnnc


23/28



( ) 111; +


24/28



ng thc xy ra khi v ch khi :

cbaCBAzyxBzCyx

BzCy::sin:sin:sin::

coscos

sinsin==

+=

=

tc zyx ,, l ba cnh ca tam gic tng ng vi ABC .

V d 1.3.2.2.

CMR Rx v ABC btk ta c :

( )CBxAx coscoscos2

11 2 +++

Li gii :

Bt ng thc cn chng minh tng ng vi :

( )( ) ( )

02

sin2

sin4

12

cos2

sin4

2sin4

2cos

2cos2

cos12coscos'

0cos22coscos2

22

22

2

2

2

2

=

=

+=

+=

++

CBA

CBA

ACBCB

ACB

ACBxx

Vy bt ng thc trn ng.

ng thc xy ra khi v ch khi :

==

=

+=

=

CBx

CB

CBx cos2cos2coscos

0

V d 1.3.2.4.

CMR trong mi ABC ta u c :2

222

2sinsinsin

++++

cbaCcaBbcAab

Li gii :

Bt ng thc cn chng minh tng ng vi :( )

( ) ( )BbccbCcAb

BbccbCcAbaa

2cos22cos2cos'

02cos22cos2cos2

222

222

+++=

+++++


25/28



( ) 02sin2sin 2 += CcAb Vy bt ng thc c chng minh xong.

V d 1.3.2.4.

Cho ABC btk. CMR :

2

3coscoscos ++ CBA

Li gii :

t ( )BACBCB

CBAk ++

=++= cos2

cos2

cos2coscoscos

01

2

cos

2

cos2

2

cos2 2 =++

+

kBABABA

Do 2

cosBA +

l nghim ca phng trnh :

012

cos22 2 =+

kxBA

x

Xt ( )122

cos' 2 +

= kBA

. tn ti nghim th :

( )

2

3coscoscos

2

31

2cos120' 2

++

CBA

kBA

k

pcm.

V d 1.3.2.5.

CMR Ryx , ta c :

( )2

3cossinsin +++ yxyx

Li gii :

t ( )2

sin212

cos2

sin2cossinsin 2yxyxyx

yxyxk+

++

=+++=

Khi 2

sinyx +

l nghim ca phng trnh :

012

cos22 2 =+

kxyx

x


26/28



( )

2

3

0121'

=

k

k

pcm.

1.3.3. nh l v hm tuyn tnh :

Xt hm ( ) baxxf += xc nh trn on [ ];

Nu( )

( )( )Rk

kf

kf

th ( ) [ ]; xkxf .

y l mtnh l kh hay. Trong mt s trng hp, khi m AM GM b tay,BCS u hng v iu kin th nh l v hm tuyn tnh mipht huy ht sc mnhca mnh. Mtpht biu ht sc ngin nhng li l li ra cho nhiu bi btngthc kh.

V d 1.3.3.1.

Cho cba ,, l nhng sthc khng m tha :

4222 =++ cba

CMR : 82

1+++ abccba

Li gii :

Ta vit li bt ng thc cn chng minh di dng :

082

11 ++

cbabc

Xt ( ) 82

11 ++

= cbabcaf vi [ ]2;0a .

Khi :

( ) ( )

( ) 08882822

0888280 22

=


27/28



V d 1.3.3.2.

CMR cba ,, khng m ta c :

( )( ) ( )3297 cbaabccbacabcab +++++++

Li gii :

tcba

cz

cba

by

cba

ax

++=

++=

++= ;; . Khi bi ton tr thnh :

Chng minh ( ) 297 +++ xyzzxyzxy vi 1=++ zyx

Khng mt tnh tng qut gi s { }zyxx ,,max= .

Xt ( ) ( ) 27977 ++= yzxyzzyxf vi

1;

3

1x

Ta c :

( )

( )


28/28


1.4.5.CBA

CBAsinsinsin8

9cotcotcot ++

1.4.6. CBAACCBBA

sinsinsin82

cos2

cos2

cos

1.4.7. CBACBA sinsinsincoscoscos1 +

1.4.8.Sbacacbcba 2

33111 4

++

++

+

1.4.9. 32++cba m

c

m

b

m

a

1.4.10.2

33++

c

m

b

m

a

m cba

1.4.11. 2plmlmlm ccbbaa ++

1.4.12.abcmcmbma cba

3111222

>++

1.4.13. ( )( )( )8

abccpbpap

1.4.14. rhhh cba 9++

1.4.15.

+

+

+

4

3sin

4

3sin

4

3sinsinsinsin

ACCBBACBA

Chuyen de Bat Dang Thuc Luong Giac (Chuong 1)

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Transcript of Chuyen de Bat Dang Thuc Luong Giac (Chuong 1)