Bdt_luong_giac_chuyen Ly Tu Trong -Can Tho

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Trng THPT chuyn L T TrngCn Th

- - - - - -

GIO TRNH

BT NG THC LNG GIC


2/106

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 thkhng chun bhnh 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 gicho rng nhng kin thc ny ly cho mt cuc hnh trnh.

Trc ht l cc bt ng thc i scbn ( 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 vdu ca tam thc bc hai, nh l vhm tuyn tnh )

Mc lc :1.1. Cc bt ng thc i scbn 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


3/106



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

+++

Bt ng thcAM GM(Arithmetic Means Geometric Means) l mt bt ng thcquen thuc v c ng dng rt rng ri.y l bt ng thc m bn c cn ghi nh rrng nht, n s l cng c hon ho cho vic chng minh cc bt ng thc. Sau y lhai cch chng minh bt ng 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

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

Cch 2 : ( ligii ca Polya )


4/106



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. Nhvyphi c t nht mt s, gi s l Aa 2

tc l 21 aAa =+= AaAaaaAaaAaaaa

2121 '' aaaa >

nn aaaaaaaa ...''... 321321


5/106



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 ABC nhn v *Nn ta lun c :

2

1

3tantantan

tantantan

++

++nnnn

CBA

CBA

Li gii :

Theo AM GMta 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 s thc tha :0coscoscoscos ++ baba

CMR : 0coscos + ba

Li gii :

Ta c :

( )( ) 1cos1cos1

0coscoscoscos

++

++

ba

baba

Theo AM GMth :


6/106



( ) ( )( )( )

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 GMth :

+

+

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 :


7/106



( )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 bt ng thcAM GMcng cc ng thc lnggic nnsc nh hng n cc bt ng thc cn hn ch. Khi ta kt hpAM GMcngBCS,Jensen hay Chebyshev th n thc s l mt v kh ng gm cho cc bt ng 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 bt ng thc thBCS (Bouniakovski Cauchy Schwartz) li l cnh tay phi ht sc c lc. Vi

AM GM ta lun phi ch iu kin cc bin l khng m, nhng i vi BCS ccbin khng b rng buc bi iu kin , ch cn l s thc cng ng. Chng minh btng thc ny cng rt ngin.

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 :


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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 ichy 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 gi vi s tip 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 bt ng thc ny b p b sung h tr cho nhau trong vic chng minh bt ngthc. Chng lng long nht th, song kim hp bch cngph thnh cng nhiubi ton kh.

Trm nghe khng bng mt thy, ta hyxt cc 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 BCSta 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


9/106



Tht vy :

( ) ( )( )

( )( )2

211

24111

2

1

22

15

2222

2222

++++

++

+++++

baba

abbaba

ab

( )( ) ( ) ( ) ( )62

1111

2222 +++

++ba

ba

Theo AM GMth ( )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 BCSth :

( ) ( )( )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.


10/106



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.


11/106



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 BCSlin 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 s thc 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.


12/106



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)(''


13/106



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 Jensenth :

==

++

+

+

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 :


14/106



Xt ( ) ( ) 22tanxxf = vi

2;0

x

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

( ) ( )( )( ) ( )( )( ) 0tantan1122tantan112222'' 2222222 >++++= xxxxxf Theo Jensenta 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 Jensenth :

+=

+=

++

+

+

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


15/106



++++

+=++

CBACBA

CBACBA

222

222

sinsinsinsinsinsin

coscoscos22sinsinsin

v2

33sinsinsin ++ CBA

2

33sinsinsin2 ++


16/106



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, C o 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


17/106


18/106



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, bt ng 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 th chng minh tt c cc kt qu nhl bi tp rn luyn.Ngoi ra ticng xin nhc vi bn c rng nhng kin thc trong phn ny khi pdng vo bi tpu cn thit c 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


19/106



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

=++

+=+=++

+=++

=++

==++


20/106



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

+


21/106



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.V chng minh ca n cn n mt s kin thc ca ton cao cp. Ta ch cn 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


22/106



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

( ) ( ) ( ) ( )

abcabab

cabafbfbac

=

cossinsin

cos:;:

pcm.

V d1.3.1.2.

Vi ba


23/106



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 khvi 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


24/106



( ) 111; +


25/106



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 bt k 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 thcxy 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

+++=

+++++


26/106



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

V d 1.3.2.4.

Cho ABC bt k. 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


27/106



( )

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 mt nh 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 bt ngthc kh.

V d 1.3.3.1.

Cho cba ,, l nhng s thc 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

=


28/106



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 :

( )

( )


29/106



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


30/106

Trng THPT chuyn L T Trng Cn Th Bt ng thc lng gicChng 2 Cc phngphp chng minh


Chng 2 :

Cc phng php chng minh

Chng minh bt ng thc i hi k nng v kinh nghim. Khng th khi khi m tam u vo chng minh khi gp mtbi bt ng thc. Ta s xem xt n thuc dngbino, nndng phngphp no chng minh. Lc vic chng minh bt ng thcmi thnh cng c.

Nhvy, c th ng u vi cc bt ng thc lng gic,bn c cn nm vngcc phngphp chng minh. s lkim ch nam cho ccbi bt ng thc. Nhngphngphp cng rt phongph v a dng : tng hp, phn tch, quy c ng, clng non gi, i bin, chn phn t cc tr Nhng theo kin ch quan ca mnh,nhng phng php tht s cn thit v thng dng s c tc gi gii thiu trongchng 2 : Cc phng php chng minh.

Mc lc :2.1. Bin i lng gic tng ng ... 322.2. S dng cc bc u cs ... 382.3. a v vector v tch v hng .. 462.4. Kt hp cc bt ng thc c in .. 482.5. Tn dng tnh n diuca hm s 572.6. Bi tp . 64


31/106



2.1. Bin i lng gic tng ng :

C th ni phngphp ny l mt phngphp xa nhTrit.N s dng cccng thc lng gic v s bin i qua li gia cc bt ng thc. c th s dng

tt phngphp ny bn c cn trang b cho mnh nhng kin thc cn thit v bin ilng gic (bn c c th tham kho thm phn 1.2. Cc ng thc,bt ng thctrong tamgic).

Thng thng th vi phng php ny, ta s a bt ng thc cn chng minh vdng bt ng thc ng hay quen thuc. Ngoi ra, ta cng c th s dng hai kt ququen thuc 1cos;1sin xx .

V d 2.1.1.

CMR :7

cos3

14sin2

14sin1

>

Li gii :

Ta c :

( )17

3cos

7

2cos

7cos

14sin2

14sin1

7

3cos

7

2cos

7

cos

14

sin2

14

5sin

14

7sin

14

3sin

14

5sin

14sin

14

3sin

14sin1

++=

++=

++=

Mt khc ta c :

( )27

cos7

3cos

7

3cos

7

2cos

7

2cos

7cos

7

2cos

7

4cos

7cos

7

5cos

7

3cos

7cos

2

1

7cos

++=

+++++=

t7

3cos;

7

2cos;

7cos

=== zyx

Khi t ( ) ( )2,1 ta c bt ng thc cn chng minh tng ng vi :

( ) ( )33 zxyzxyzyx ++>++

m 0,, >zyx nn :

( ) ( ) ( ) ( ) ( )403 222 >++ xzzyyx


32/106


33/106



V d 2.1.4.

Cho ( )Zkk +

2

,, l bagc tha 1sinsinsin 222 =++ . CMR :

222

2

tantantan213

tantantantantantan

++

Li gii :

Ta c :

222222222

222

222

222

tantantan21tantantantantantan

2tan1

1

tan1

1

tan1

1

2coscoscos

1sinsinsin

=++

=+

++

++

=++

=++

Khi bt ng thc cn chng minh tng ng vi :

( ) ( ) ( ) 0tantantantantantantantantantantantan

tantantantantantan3

tantantantantantan

222

222222

2

++

++

++

pcm.

ng thc xy ra

tantantan

tantantantan

tantantantan

tantantantan

==

=

=

=

V d 2.1.5.

CMR trong ABC bt k ta c :

++++

2tan

2tan

2tan3

2cot

2cot

2cot

CBACBA

Li gii :

Ta c :

2cot

2cot

2cot

2cot

2cot

2cot

CBACBA=++

t2

cot;2

cot;2

cotC

zB

yA

x === th

=++

>

xyzzyx

zyx 0,,



34/106



( ) ( )

( ) ( )( ) ( ) ( ) 0

3

3

1113

222

2

++

++++

++++

++++

xzzyyx

zxyzxyzyx

xyz

zxyzxyzyx

zyxzyx

pcm.ng thc xy ra CBA cotcotcot ==

CBA ==

ABC u.

V d 2.1.6.

CMR :xxx cos2

2sin31

sin31

+

+

+

Li gii :

V 1sin1 x v 1cos x nn :0sin3;0sin3 >>+ xx v 0cos2 >+

Khi bt ng thc cn chng minh tng ng vi :( ) ( )

( )

( )( ) 02cos1cos

04cos6cos2

cos1218cos612

sin92cos26

2

2

2

+

+

+

xx

xx

xx

xx

do 1cos x nn bt ng thc cui cng lun ng pcm.

V d 2.1.7.

CMR2

;3


35/106


36/106



V d 2.1.9.

Cho ABC khng vung. CMR :

ACCBBACBACBA 222222222222 tantantantantantan9tantantan5tantantan3 +++++

Li gii :

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

( )

( ) ( )

( )

( )( ) ( ) 0sincoscos2

01coscos4cos4

01cos4coscos2

01cos42cos2cos2

4

3

cos2

2cos1

2

2cos1

4

3coscoscos

coscoscos

1

coscos

1

coscos

1

coscos

1

coscoscos

4

coscoscos

183

cos

1

cos

1

cos

141

cos

11

cos

11

cos

14

tan1tan1tan18tantantan4tantantan4

22

2

2

2

2

222

222222222222

222222222

222222222

+

+

+++

+++

+

+

+

+

++

++

++

+++++

BABAC

BACC

CBABA

CBA

C

BA

CBA

CBAACCBBACBA

CBACBACBA

CBACBACBA

pcm.

V d sau y, theo kin ch quan ca tcgi, th ligii ca nxng ng l bcthy v bin i lng gic. Nhng bin i tht s lt lo kt hp cng bt ng thcmt cch hp l ng ch mang n cho chng ta mt bi ton tht s c sc !!!

V d 2.1.10.

Cho na ng trn bn knh R , C l mt im ty trn na ng trn. Trong haihnh qut ni tip hai ng trn, gi M v N l hai tip im ca hai ng trn vi

ng knh ca na ng trn cho. CMR : 122 RMN

Li gii :

Gi 21 ,OO l tm ca hai ng trn. t 2=CON (nhvy 20


37/106



NM O

O1O2

C

Vy :

cottancot2

cot 2121 RRRRONMOMN +=+

=+=

Trong vung MOO1 c :

( )

( )

cos1

coscoscos1

cos2sin

11

111

+==+

=

=

RRRR

RROOR

Tng t :

( )

sin1

sinsinsin 2222

+===

RRRROOR

Do :

( )( )

1cossin

2

2cos

2sin

2cos

1

2cos2.

2cos

2sin

2cos

2sin

2cos2

cos1sin1

1cossin

sin1cos

cos1sin

sin

cos

sin1

sin

cos

sin

cos1

cos

2

2

++=

+

=

+

+

=

++

++=

++

+=

+

++

=

R

R

R

R

RR

RRMN

m ( )=+

+ 122

12

22

42cossin R

RMN

pcm.

ng thc xy ra MNOC=4

.

2.2. S dng cc bc u cs :

Cc bc u cs m tcgi mun nhc n y lphn 1.2. Cc ng thc, btng thc trong tam gic. Tas a cc bt ng thc cn chng minh v cc bt ngthc cbn bng cch bin i v s dng cc ng thc cbn.Ngoi ra, khi tham giacck thi, tcgi khuyn bn c nn chng minh cc ng thc, bt ng thc cbns dng nhmt b cho bi ton.


38/106



C1

C

B1

B

A1

A

V d 2.2.1.

Cho ABC .ng phngic trong ccgc CBA ,, ct ng trn ngoi tip ABC

ln lt ti 111 ,, CBA . CMR :

111 CBAABCSS

Li gii :

Gi R l bn knh ng trn ngoi tip ABC th n cng l bn knh ng trnngoi tip 111 CBA .

Bt ng thc cn chng minh tng ng vi :( )1sinsinsin2sinsinsin2 111

22 CBARCBAR

Do 2;2;2111

BA

C

AC

B

CB

A

+

=

+

=

+

= nn :

( )

( )22

cos2

cos2

cos2

cos2

cos2

cos2

sin2

sin2

sin8

2sin

2sin

2sinsinsinsin1

CBACBACBA

BAACCBCBA

+++

V 02

cos2

cos2

cos >CBA

nn :

( ) 8

1

2sin

2sin

2sin2

CBApcm.

ng thc xy ra ABC

u.

V d 2.2.2.

CMR trong mi tamgic ta u c :

2sin

2sin

2sin4

4

7sinsinsinsinsinsin

CBAACCBBA +++

Li gii :

Ta c :2

sin2

sin2

sin41coscoscos CBACBA +=++

Bt ng thc cho tng ng vi :

( )1coscoscos4

3sinsinsinsinsinsin CBAACCBBA +++++

m :


39/106



BABAC

ACACB

CBCBA

coscossinsincos

coscossinsincos

coscossinsincos

=

=

=

nn :

( ) ( )243

coscoscoscoscoscos1 ++ ACCBBA

Tht vy hin nhin ta c :

( ) ( )3coscoscos3

1coscoscoscoscoscos

2CBAACCBBA ++++

Mt khc ta c :2

3coscoscos ++ CBA

( )3 ng ( )2 ng pcm.ng thc xy ra khi v ch khi ABC u.

V d 2.2.3.

Cho ABC bt k. CMR :

1coscos4cos21

1

coscos4cos21

1

coscos4cos21

1

+++

+++

++ ACCCBBBAA

Li gii :

t v tri bt ng thc cn chng minh l T.Theo AM GMta c :

( ) ( )[ ] ( )19coscoscoscoscoscos4coscoscos23 ++++++

ACCBBACBAT m :

2

3coscoscos ++ CBA

v hin nhin :( )

4

3

3

coscoscoscoscoscoscoscoscos

2

++

++CBA

ACCBBA

( ) ( ) ( )29coscoscoscoscoscos4coscoscos23 ++++++ ACCBBACBA

T ( ) ( )2,1 suy ra 1T pcm.

V d 2.2.4.

CMR vi mi ABC bt k, ta c :

( ) ( ) ( )222222 34 accbbaScba +++++

Li gii :



40/106


41/106



+

2tansin

2tansin

2

1

2sin

2sin2

AB

BA

BA

Tng t ta c :

+

+

2tansin

2tansin

2

1

2sin

2sin2

2tansin

2tansin

2

1

2sin

2sin2

CA

AC

AC

BC

CB

CB

T suy ra :

( ) ( ) ( )

+++++

++

BAC

ACB

CBA

ACCBBA

sinsin2

tansinsin2

tansinsin2

tan2

1

2sin

2sin

2sin

2sin

2sin

2sin2

++++

2

sin

2

sin

2

sin

2

sin

2

sin

2

sin2coscoscosACCBBA

CBA

Khi :

( )

( ) ( ) ( )4

1coscoscos

4

11coscoscos

4

1coscoscos

2

1

1coscoscos4

1

2sin

2sin

2sin

2sin

2sin

2sin

=++=++++

++++

CBACBACBA

CBAACCBBA

m2

3coscoscos ++ CBA

( )8

51coscoscos

4

1

2sin

2sin

2sin

2sin

2sin

2sin ++++ CBA

ACCBBA

( )2 ngpcm.

V d 2.2.6.

Cho ABC bt k. CMR :

2tan

2tan

2tan

cotcotcot

2223

222

CBA

cba

CBA

cba

++

++

Li gii :

Ta c :

SCBA

cba4

cotcotcot

222

=++

++

nn bt ng thc cho tng ng vi :


42/106



( )1

2tan

2tan

2tan

64222

3

CBA

cbaS

Mt khc ta cng c :

2sin4

cos22cos2

22

2222

Abca

AbcbcaAbccba

+=

SAbcA

Abc

A

a4sin2

2tan

2sin4

2tan

22

==

Tng t ta cng c :

SC

cS

B

b4

2

tan

;4

2

tan

22

( )1 ng pcm.

V d 2.2.7.

CMR trong mi tamgic ta c :( ) ( ) ( ) 3cos1cos1cos1 ++++++++ CabbaBcaacAbccb

Li gii :

Ta c v tri ca bt ng thc cn chng minh bng :( ) ( ) ( ) ( )[ ] ( )BcaAbcCabCbaBacAcbCBA coscoscoscoscoscoscoscoscos ++++++++++

t :

( ) ( ) ( )

BcaAbcCabR

CbaBacAcbQ

CBAP

coscoscos

coscoscos

coscoscos

++=

+++++=

++=

D thy2

3P

Mt khc ta c :( ) ( ) aARCBRBCCBRBcCb ==+=+=+ sin2sin2cossincossin2coscos

Tng t :

cbaQ

cAbBa

bCaAc

++=

=+

=+

coscos

coscos

V ta li c :


43/106


44/106



22

3

8

23

8

++

++

+

R

S

ac

caca

cb

bcbc

ba

abab

r

S

Li gii :

Theo AM GM ta c :

2

cabcab

ac

caca

cb

bcbc

ba

abab ++

++

++

+

Do( )

623

822

cba

r

SprS

++=

=

Li c :

( )62

2cbacabcab ++

++

++

++

+

ac

caca

cb

bcbc

ba

abab

r

S2

23

8

v tri c chng minh xong.Ta c :

( )

33

2

33sinsinsin

sinsinsin2

Rcba

CBA

CBARcba

++

++

++=++

Theo AM GM ta c :

( )( ) ( )( ) ( )( )8

2 abcpapcpcpbpbpappS =

( ) ( ) ( )accbba

abc

cba

abc

cba

abcp

R

S

+++++=

++=

++

9

2

9

33

8

3

8

3

82

2

Mt ln na theo AM GM ta c :

( ) ( ) ( ) ( )( )( ) accaca

cb

bcbc

ba

abab

accbba

abc

accbba

abc

++

++

+

+++

+++++ 3.3

99

v phi chng minh xongBt ng thc c chng minh hon ton.

V d 2.2.10.

Cho ABC bt k. CMR :4

2

8

2

8

2

8

3

6

2cos

2cos

2cos

++

R

abc

C

c

B

b

A

a


45/106



Li gii :

p dng BCS ta c :

( )

2cos2cos2cos2cos2cos2cos

222

2444

2

8

2

8

2

8

CBA

cba

C

c

B

b

A

a

++

++++

m :

( )224

222

16

4

9

2cos

2cos

2cos

SR

abc

CBA

=

++

V th ta ch cn chng minh : 2444 16Scba ++ Trc ht ra c : ( ) ( )1444 cbaabccba ++++

Tht vy : ( ) ( ) ( ) ( ) 01 222222 ++ abcccabbbcaa

( ) ( ) ( ) ( ) ( ) ( ) 0222222222 ++++++++ babacacacbcbcba (ng!)Mt khc ta cng c :

( )( )( ) ( )( )( )( ) ( )21616 2 bacacbcbacbacpbpappS +++++==

T ( ) ( )2,1 th suy ra taphi chng minh : ( )( )( ) ( )3bacacbcbaabc +++ t :

bacz

acby

cbax

+=

+=

+=

v cba ,, l ba cnh ca mt tam gic nn 0,, >zyx Khi theo AM GMth :

( )( )( ) ( )( )( )( )( )( )bacacbcbaxyz

zxyzxyxzzyyxabc +++==

+++=

8

222

8

( )3 ng pcm.

2.3 a v vector v tch v hng :

Phng php ny lun a ra cho bn c nhng li gii bt ng v th v. N ctrng cho s kt hp hon gia i s v hnh hc. Nhng tnh cht ca vector li mang

n ligii thtsngsa v p mt. Nhng s lng cc bi ton ca phngphp nykhng nhiu.

V d 2.3.1.


46/106



A

BC

e

e

e

1

2

3

O

A

B C

CMR trong mi tamgic ta c :

2

3coscoscos ++ CBA

Li gii :

Ly cc vector n v 321 ,, eee ln lt trn cc cnh CABCAB ,, .

Hin nhin ta c :

( )( ) ( ) ( )

( )

2

3coscoscos

0coscoscos23

0,cos2,cos2,cos23

0

133221

2

321

++

++

+++

++

CBA

CBA

eeeeee

eee

pcm.

V d 2.3.2.

Cho ABC nhn. CMR :

2

32cos2cos2cos ++ CBA

Li gii :

Gi O, G ln lt l tm ng trn ngoi tip v trng tm ABC .

Ta c : OGOCOBOA 3=++ Hin nhin :

( )( ) ( ) ( )[ ]

( )

2

32cos2cos2cos

02cos2cos2cos23

0,cos,cos,cos23

0

22

22

2

++

+++

+++

++

CBA

BACRR

OAOCOCOBOBOARR

OCOBOA

pcm.

ng thc xy ra ABCGOOGOCOBOA ==++ 00 u.

V d 2.3.3.


47/106



O

A

B C

Cho ABC nhn. CMR Rzyx ,, ta c :

( )2222

12cos2cos2cos zyxCxyBzxAyz ++++

Li gii :

Gi O l tm ng trn ngoi tip ABC .Ta c :

( )

( )222

222

222

2

2

12cos2cos2cos

02cos22cos22cos2

0.2.2.2

0

zyxCxyBzxAyz

BzxAyzCxyzyx

OAOCzxOCOByzOBOAxyzyx

OCzOByOAx

++++

+++++

+++++

++

pcm.

2.4. Kt hp cc bt ng thc c in :

V ni dung cng nhcch thc s dng cc bt ng thc chng ta bn chng1:Cc bc u cs. V th phn ny, tas khng nhc li m xt thm mt s vdphc tp hn, th v hn.

V d 2.4.1.

CMR ABC ta c :

2

39

2cot

2cot

2cot

2sin

2sin

2sin

++

++

CBACBA

Li gii :

Theo AM GM ta c :

3

2

sin

2

sin

2

sin

3

2sin

2sin

2sin

CBA

CBA

++

Mt khc :

2sin

2sin

2sin

2cos

2cos

2cos

2cot

2cot

2cot

2cot

2cot

2cot

CBA

CBA

CBACBA==++


48/106


49/106



( )

CBA

CCBBAA

CBA

CCBBAA

CBA

CBA

coscoscos2

cossincossincossin

2

3

coscoscos2

cossincossincossin

coscoscos

2sin2sin2sin4

1

3

++=

++

=

Suy ra :

( )( )

( )1tantantan2

9

coscoscos

cossincossincossincoscoscos

2

9tantantancoscoscos

3

3

CBA

CBA

CCBBAACBACBACBA

=

++++

Mt khc : 33tantantan CBA

( )22

3933

2

9tantantan

2

9 33= CBA

T ( )1 v ( )2 suy ra :

( )( )2

39tantantancoscoscos ++++ CBACBA

pcm.

V d 2.4.3.

Cho ABC ty. CMR :

34

2tan

1

2tan

2tan

1

2tan

2tan

1

2tan

++

++

+C

C

B

B

A

A

Li gii :

Xt ( )

=

2;0tan

xxxf

Khi : ( ) =xf ''

Theo Jensen th : ( )132

tan2

tan2

tan ++CBA

Xt ( )

= 2;0cot

xxxg

V ( ) ( )

>+=

2;00cotcot12'' 2

xxxxg

Theo Jensen th : ( )2332

cot2

cot2

cot ++CBA

Vy ( ) ( )+ 21 pcm.


50/106



V d 2.4.4.

CMR trong mi tamgic ta c :3

3

21

sin

11

sin

11

sin

11

+

+

+

+

CBA

Li gii :

Ta s dng b sau :B : Cho 0,, >zyx v Szyx ++ th :

( )12

11

11

11

1

3

+

+

+

+

Szyx

Chng minh b :Ta c :

( ) ( )2111111111 xyzzxyzxyzyxVT +

+++

+++=

Theo AM GM ta c :

( )399111

Szyxzyx

++++

Du bng xy ra trong ( )3

3S

zyx ===

Tip tc theo AM GM th :33 xyzzyxS ++

( )4271

27 3

3

Sxyzxyz

S

Du bng trong ( )4 xy ra3

Szyx ===

Vn theo AM GM ta li c :

( )51

3111

3

2

++

xyzzxyzxy

Du bng trong ( )5 xy ra3

Szyx ===

T ( ) ( )54 suy ra :

( )627111

2Szxyzxy++

Du bng trong ( )6 xy ra ng thi c du bng trong ( ) ( )3

54S

zyx ===

T ( )( )( )( )6432 ta c :


51/106



( )3

32

31

2727911

+=+++

SSSSVT

B c chng minh. Du bng xy ra ng thi c du bng trong ( )( )( )643

3

Szyx ===

p dng vi 0sin,0sin,0sin >=>=>= CzByAx

m ta c2

33sinsinsin ++ CBA vy y

2

33=S

Theo b suy ra ngay :3

3

21

sin

11

sin

11

sin

11

+

+

+

+

CBA

Du bng xy ra2

3sinsinsin === CBA

ABC

u.

V d 2.4.5.

CMR trong mi tamgic ta c :

3plll cba ++

Li gii :

Ta c : ( ) ( ) ( )1222cos2 appcb

bc

bc

app

cb

bc

cb

A

bcla

+=

+=

+=

Theo AM GM ta c 12

+ cb

bcnn t ( )1 suy ra :

( ) ( )2appla

Du bng trong ( )2 xy ra cb = Hon ton tng t ta c :

( ) ( )

( ) ( )4

3

cppl

bppl

c

b

Du bng trong ( ) ( )43 tng ng xy ra cba ==

T ( )( ) ( )432 suy ra :

( ) ( )5cpbpapplll cba ++++ Du bng trong ( )5 xy ra ng thi c du bng trong ( )( ) ( ) cba ==432p dng BCS ta c :


52/106



( ) ( )( )63

332

pcpbpap

cbapcpbpap

++

++

Du bng trong ( )6 xy ra cba ==

T ( ) ( )65 ta c : ( )73plll cba ++

ng thc trong ( )7 xy ra ng thi c du bng trong ( ) ( ) cba ==65ABC u.

V d 2.4.6.

Cho ABC bt k. CMR :

R

r

abc

cba 24

333

++

Li gii :

Ta c : ( )( )( )cpbpappprR

abcS ===

4

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

( )( )( )abc

abccbacaacbccbabba

abc

cbabacacb

abc

cpbpap

pabc

cpbpapp

pabc

S

R

r

2

222222882

333222222

2

+++++=

+++=

=

==

abccba

ca

ac

bc

cb

ab

ba

abccba

Rr

333333

624 ++

++++++

++=

pcm.

V d 2.4.7.

Cho ABC nhn. CMR :

abcbA

a

C

ca

C

c

B

bc

B

b

A

a27

coscoscoscoscoscos

+

+

+

Li gii :


CBABA

A

C

CA

C

C

B

BC

B

B

A

Asinsinsin27sin

cos

sin

cos

sinsin

cos

sin

cos

sinsin

cos

sin

cos

sin

+

+

+


53/106



27coscos

coscos1

coscos

coscos1

coscos

coscos1

sinsinsin27sincoscos

sinsin

coscos

sinsin

coscos

sin

AC

AC

CB

CB

BA

BA

CBABAC

BA

CB

AC

BA

C

t

+

=

+

=

+

=


54/106



Li gii :

Bt ng thc cn chng minh tng dng vi :

( )

( ) ( )cba

abccbacba

cba

abccbacba

+++++++

+++

++++

72935

2

435

36

2222

2

222

Theo BCS th : ( ) ( )2222 3 cbacba ++++ ( ) ( ) ( )1279 2222 cbacba ++++

Li c :

++

++

3 222222

3

3

3

cbacba

abccba

( )( )( )( )

( ) ( )2728

7289

222

222

222

cba

abccba

abccbacbaabccbacba

++++

++++

++++

Ly ( )1 cng ( )2 ta c :

( ) ( ) ( )

( ) ( )cba

abccbacba

cba

abccbacbacba

+++++++

++++++++++

72935

729827

2222

2222222

pcm.

V d 2.4.9.

CMR trong ABC ta c :

6

2sin

2cos

2sin

2cos

2sin

2cos

+

+

C

BA

B

AC

A

CB

Li gii :

Theo AM GM ta c :

( )1

2sin

2cos

2sin

2cos

2sin

2cos

3

2sin

2cos

2sin

2cos

2sin

2cos

3C

BA

B

AC

A

CB

C

BA

B

AC

A

CB

+

+


55/106



m :

( )( )( )CBA

BAACCB

CC

BABA

BB

ACAC

AA

CBCB

C

BA

B

AC

A

CB

sinsinsinsinsinsinsinsinsin

2sin

2cos2

2cos

2sin2

2sin

2cos2

2cos

2sin2

2sin

2cos2

2cos

2sin2

2sin

2cos

2sin

2cos

2sin

2cos

+++=

+

+

+

=

Li theo AM GM ta c :

+

+

+

ACAC

CBCB

BABA

sinsin2sinsin

sinsin2sinsin

sinsin2sinsin

( )( )( )

( )( )( )( )28

sinsinsin

sinsinsinsinsinsin

sinsinsin8sinsinsinsinsinsin

+++

+++

CBA

BAACCB

CBABAACCB

T ( )( )21 suy ra :

683

2sin

2cos

2sin

2cos

2sin

2cos

3=

+

+

C

BA

B

AC

A

CB

pcm.

V d 2.4.10.

CMR trong mi ABC ta c : 29sinsinsinsinsinsin

++

R

rACCBBA

Li gii :


2

2

2

36

9222222

9sinsinsinsinsinsin

rcabcab

raccbba

rACRCBRBAR

++

++

++

Theo cng thc hnh chiu :

+=

+=

+=

a

BArc

a

ACrb

a

CBra cot

2cot;cot

2cot;cot

2cot


56/106



+

++

+

+

++

+

+=++

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2

22

CBBAr

BAACr

ACCBrcabcab

Theo AM GM ta c :

( )1cotcotcot42

cot2

cot22

cot2

cot22

cot2

cot2

cot2

cot 2 BACACCBACCB

=

+

+

Tng t :

( )

( )3cotcotcot42

cot2

cot2

cot2

cot

2cotcotcot42

cot2

cot2

cot2

cot

2

2

ACBCBBA

CBABAAC

+

+

+

+

T ( )( )( )321 suy ra :

( )42

cot2

cot2

cot122

cot2

cot2

cot2

cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

3222 CBABAAC

BAACBAAC

+

++

+

+

++

+

+

Mt khc ta c : ( )5272

cot2

cot2

cot332

cot2

cot2

cot 222 CBACBA

T ( ) ( )54 suy ra : ( )6363.122

cot2

cot2

cot123 222 =CBA

T ( ) ( )64 suy ra pcm.

2.5. Tn dng tnh n iu ca hm s :

Chng ny khi c th bn c cn c kin thc cbn v o hm, khost hm sca chng trnh 12 THPT. Phngphp ny thc s c hiu qu trong cc bi bt ngthc lnggic. c th s dng tt phngphp ny th bn c cn n nhng kinhnghimgii ton cc phngphp nu cc phn trc.

V d 2.5.1.

CMR :

xx

2sin > vi

2;0

x

Li gii :


57/106



Xt ( )

2sin=

x

xxf vi

2;0

x

( )2

sincos'

x

xxxxf

=

Xt ( ) xxxxg sincos = vi

2;0

x

( ) ( )xgxxxxg

vi

2;0

Li gii :


( )

( ) 0cossin

cossin

3

1

3

1

>

>

xx

x

x

Xt ( ) ( ) xxxxf =

3

1

cossin vi

2;0 x

Ta c : ( ) ( ) ( ) 1cossin3

1cos' 3

42

3

2

=

xxxxf

( ) ( ) ( ) ( )

>+=

2;00cossin

9

4sin1cos

3

2'' 4

73

3

1 xxxxxxf

( )xf ' ng bin trong khong ( ) ( ) 00'' => fxf

( )xf cng ng bin trong khong ( ) ( ) => 00fxf pcm.

V d 2.5.3.

CMR nu a l gc nhn hay 0=a th ta c :1tansin 222 ++ aaa

Li gii :


58/106



p dng AM GM cho hai s dng asin2 v atan2 ta c :aaaaaa tansintansintansin

2222222 +

=+

Nhvy ta ch cn chng minh : aaa 2tansin >+ vi2

0

+=

+=+=

2;00

cos

cos1cos1cos1

cos

1cos2cos2

cos

1cos'

22

23

2

x

x

xxx

x

xx

xxxf

( )xf ng bin trn khong ( ) ( )0faf > vi aaaa 2tansin2

;0 >+

12tansin22222

++= aaaa

1tansin 222 ++ aaa (khi 0=a ta c du ng thc xy ra).

V d 2.5.4.

CMR trong mi tamgic ta u c :

( ) CBACBABABABA coscoscoscoscoscos12

13coscoscoscoscoscos1 ++++++

Li gii :

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

6

131coscoscoscoscoscos2coscoscos21 ++++++

( ) ( CBABABABACBA coscoscos6

131coscoscoscoscoscos2coscoscos 222 ++++++++

( ) ( )CBACBA coscoscos6

131coscoscos

2+++++

6

13

coscoscos

1coscoscos

+++++

CBACBA

t

2

31coscoscos =

2

3;10

11'

2ng bin trn khong .

( ) =

6

13

2

3fxf pcm.


59/106



V d 2.5.5.

Cho ABC c chu vi bng 3. CMR :

( )2

222

4

13sinsinsin8sinsinsin3

RCBARCBA +++

Li gii :

Bt ng thc cn chng minh tng ng vi :( )( )( ) 13sin2sin2sin24sin4.3sin4.3sin4.3 222222 +++ CRBRARCRBRAR

134333222

+++ abccba

Do vai tr ca cba ,, l nhnhau nn ta c thgis cba

Theo githit :2

3133 >+=++ ccccbacba

Ta bin i :

( )( )[ ]( )

( ) ( )

( ) ( )cabcc

cabcc

ababccc

abccabba

abccba

abccbaT

232333

322333

64333

4323

433

4333

22

22

22

22

222

222

+=

++=

++=

+++=

+++=

+++=

v 0230322

3>


60/106



Li gii :

Ta c :

( )( )

( )

( )( )

( )

( )( )

( )

p

cp

p

bp

p

apCBA

cpp

bpapC

bpp

apcpB

app

cpbpA

=

=

=

=

2tan

2tan

2tan

2tan

2tan

2tan

v( )( )( )

p

cp

p

bp

p

ap

p

cpbpapp

p

S

S

r

=

==

22

2

Do :2

tan2

tan2

tan2 CBA

S

r=

Mt khc :

( ) ( )

( )

( )

2cot

2cot

2cot

2cos

2sin

2sin

2sin

2cos

2cos

2cos

2cos

2tansinsinsin2

sinsinsin2

2tan

2tan2

CBA

A

A

CBA

CBA

AACBR

CBAR

Aacb

cba

Aap

cba

r

p

==

+

++=

+

++=

++=


33

28

2cot

2cot

2cot

2cot

2cot

2cot

1

33

28

2cot

2cot

2cot

2tan

2tan

2tan

+

+

CBA

CBA

CBACBA

t 332

cot2

cot2

cot = tCBA

t

Xt ( )t

ttf1

+= vi 33t

( ) 3301

1'2

>= tt

tf

( ) ( ) =+==33

28

33

13333min ftf pcm.

V d2.5.7.


61/106



CMR vi mi ABC ta c :

( )( )( ) 233

38222 eRcRbRaR


62/106


63/106



V d2.5.10.

CMR :20

720sin

3

1 0


64/106



2.6.1. ( )2

5cos2cos2cos3 + BCA

2.6.2. 42cos322cos22cos3 ++ CBA

2.6.3. ( ) 542cos532cos2cos15 ++++ CBA

2.6.4. 342

tan2

tan2

tan ++ CBA vi ABC c mt gc 32

2.6.5.2222 4

1111

rcba++

2.6.6.cba r

c

r

b

r

a

r

abc 333++

2.6.7.( )( )( )

23

k Gii: Trc ht ta chng minh :B 1: 0, > yx v 01 > k th :

( ) ( ) ( )Hyxyx kkkk ++ 12

Chng minh:( ) ( ) ( ) ( ) 0121121 11 ++=

+

+

kkk

k

kk

k

aaafy

x

y

xH vi 0>= a

y

x

V ( ) ( ) ( ) 021' 11 =+= kk aakaf 1= a hoc 1=k . Vi 1=k th ( )H l ng thcng.Do 0>a v 01 >> k th ta c :

( ) 00 > aaf v 01 >> k


78/106

Trung THPT chuyn L T Trng Cn Th Bt ng thc lng gicChng 4Mt s chuyn bi vit hay,th v

lin quan n bt ng thc v lnggic


( )H c chng minh.Tr libi ton 2 :T h ( )1 ta c :

+

+

k

b

k

ck

k

bck

a a

cd

a

bd

a

cd

a

bdR 12

( p dng b ( )H via

cdy

a

bdx bc == ; )

Tng t :

+

+

k

a

k

bkk

c

k

a

k

ckk

b

c

bd

c

adR

b

cd

b

adR

1

1

2

2

( )kck

b

k

a

k

kkk

c

kkk

b

kkk

a

kk

c

k

b

k

a

ddd

a

b

b

ad

a

c

c

ad

b

c

c

bdRRR

++

+

+

+

+

+

++

2

2 1

pcm.ng thc xy ra khi ABC u v M l tm tam gic. p dng ( )E ta chng minhcbi ton sau :Bi ton 3: Chng minh rng :

( )3111

2111

++++

cbacba RRRddd

Gii: Thc hinphp nghch o tm M, phng tch n v ta c :

=

=

=

c

b

a

RMC

RMB

RMA

1*

1*

1*

v

=

=

=

c

b

a

dMC

dMB

dMA

1''

1''

1''

p dng ( )E trong '''''' CBA :

( )

++++

++++

cbacba RRRddd

MCMBMAMCMBMA111

2111

***2''''''

pcm.M rng kt qu ny ta c bi ton sau :Bi ton 4: Chng minh rng :

( )42 kck

b

k

a

k

c

k

b

k

a

kRRRddd ++++


79/106




vi 10 > k Hng dn cch gii : Ta thy ( )4 d dng c chng minh nh p dng ( )2 trongphp bin hnh nghch o tm M, phng tch n v. ng thc xy ra khi ABC uv M l tm tam gic.By gi vi 1>k th t h ( )1 ta thu c ngay :Bi ton 5: Chng minh rng :

( )( )52 222222 cbacba dddRRR ++>++ Xutpht t bi ton ny, ta thu c nhng kt qu tng qut sau :Bi ton 6: Chng minh rng :

( )( )62 kck

b

k

a

k

c

k

b

k

a dddRRR ++>++

vi 1>k Gii: Chng ta cng chng minh mt b :B 2: 0, > yx v 1>k th :

( ) ( )Gyxyx kkk ++

Chng minh:

( ) ( ) ( ) 01111 >+=+>

+ k

k

k

kk

aaagy

x

y

xG (t 0>= a

y

x)

V ( ) ( ) 1;001' 11 >>>+= kaaakag kk

( ) 1;00 >>> kaag

( )G c chng minh xong.

S dng b ( )G vobi ton ( )6 :T h ( )1 :

k

b

k

c

k

bck

aa

cd

a

bd

a

cd

a

bd

R

+

>

+ (t a

cd

ya

bd

xbc

== ; )

Tng t:

k

a

k

bk

c

k

a

k

ck

b

c

bd

c

adR

b

cd

b

adR

+

>

+

>

( )k

c

k

b

k

a

kk

k

c

kk

k

b

kk

k

a

k

c

k

b

k

a

ddd

a

b

b

ad

a

c

c

ad

b

c

c

bdRRR

++

+

+

+

+

+

>++

2

pcm.Bi ton 7: Chng minh rng :

( ) ( )72 kak

a

k

a

k

a

k

a

k

a RRRddd ++>++

vi 1


80/106




Hng dn cch gii : Ta thy ( )7 cng c chng minh d dng nh p dng ( )6 trongphp bin hnh nghch o tm M, phng tch n v. ng thc khng th xy ratrong ( )6 v ( )7 .Xt v quan h gia ( )cba RRR ,, vi ( )cba ddd ,, ngoi bt ng thc ( )E v nhng m

rng ca n, chng ta cn gp mt s bt ng thc rt hay sau y. Vic chng minhchng xin dnh chobn c :

( )( )( )

( )( )( )ccbbccaabbaacba

cbcabacba

c

ba

b

ca

a

cb

cbacba

dRdRdRdRdRdRRRR

ddddddRRR

R

dd

R

dd

R

dd

dddRRR

+++

+++

+

++

++

222)4

)3

3)2

8)1


81/106




ng dng ca i s vo vic pht hin v chngminh bt ng thc trong tam gic

L Ngc Anh

(HS chuyn ton kha 2005 2008Trng THPT chuyn L T Trng, Cn Th)

1/ Chng ta i tbi ton i ssau:Vi

x

0, 0, 0, 0,2222

ta lun c:

x x 2x< tg < < sinx < x

2 2 .

Chng minh:Ta chng minh 2 bt ng thc:2

sinx

x

> v2

2

x xtg

< .

t1

( ) sinf x xx

= l hm sxc nh v lin tc trong 0,2

.

Ta c:2

os x- sin x'( )

xcf x

x= . t ( ) os x- sin xg x xc= trong 0,

2

khi

( ) ( )' sin 0g x x x g x= nghch bin trong on 0,2

nn ( ) ( )0g x g< =0 vi

0,2

x

. Do ( )' 0f x < vi 0,

2x

suy ra ( )2

2f x f

> =

hay

2sin

xx

>

vi 0,2

x

.

t ( ) 1h x tgxx

= xc nh v lin tc trn 0,2

.

Ta c ( )2 2

sin' 0

2 os2

x xh x

xx c

= > 0,

2x

nn hm s ( )h x ng bin, do

( )2 2

xh x h

< =

hay

2

2

x xtg

< vi 0,

2x

.

Cn 2 bt ng thc2 2

x xtg > v sinx x< dnh cho bn c tchng minh.

By gimi l phn ng ch :Xt ABC: BC = a , BC = b , AC = b . GiA, B, Cl ln cc gc bng radian;

r, R, p, Sln lt l bn knh ng trn ni tip, bn knh ng trn ngoi tip, nachu vi v din tch tam gic; la, ha, ma, ra,tng ng l di ng phn gic, ngcao, ng trung tuyn v bn knh ng trn bng tip ng vi nhA...

Bi ton 1:Chng minh rng trong tam gic ABC nhn ta lun c:

2 2 2os os os4

p pAc x Bc B Cc C

R R

< + + <


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Nhn xt:

Tnh l hm s sin quen thuc trong tam gic ta c: sin sin sinp

A B BR

+ + = v

bi ton i sta ddng a ra bin i sau 2 2 24

os 2 os sin os2

AAc A tg c A A Ac A

< = < , t

a n li gii nhsau.Li gii:

Ta c: 2 2 24

os 2 os sin os2

AAc A tg c A A Ac A

< = < 2os sin

pAc A A

R< =

v 2 24

os sin os4

p pAc A A Ac A

R R

> = > . Ty suy ra pcm.

Trong mt tam gic ta c nhn xt sau: 12 2 2 2 2 2

A B B C C Atg tg tg tg tg tg+ + = kt hp

vi2

2

x xtg

< nn ta c

2 2 2 2 2 21

2 2 2 2 2 2

A B B C C A A B B C C Atg tg tg tg tg tg

+ + > + + =

2

. . .4

A B B C C A

+ + > (1). Mt khc2 2

x xtg > nn ta cng d dng c

12 2 2 2 2 2 2 2 2 2 2 2

A B B C C A A B B C C Atg tg tg tg tg tg+ + < + + = t y ta li c

. . . 4A B B C C A+ + < (2). T(1) v (2) ta c bi ton mi.Bi ton 2:Chng minh rng trong tam gic ABC nhn ta lun c:

2

. . . 44

A B B C C A

< + + <

Lu :Khi dng cch ny sng to bi ton mi th ton l ABC phi l nhn

v trong bi ton i sth 0, 2x

. Li gii bi ton tng tnhnhn xt trn.

Mt khc, p dng bt ng thc( )

2

3

a b cab bc ca

+ ++ + th ta c ngay

( )2 2

. . .3 3

A B CA B B C C A

+ ++ + = . Ty ta c bi ton cht hn v p hn:

2 2

. . .4 3

A B B C C A

+ +

By gita thi tcng thc la, ha, ma, ra tm ra cc cng thc mi.

Trong ABC ta lun c: 2 sin sin sin2 2a aA A

S bc A cl bl= = +


83/106




1 1 1 1

A 2 22 os2

a

b c b c

l bc b cbcc

+ + = > = +

1 1 1 1 1 1 1 1 1 1

2 sin sin sina b cl l l a b c R A B C

+ + > + + > + +

1 1 1 1 1 1 1

2a b c

l l l R A B C

+ + > + +

.

Nhvy chng ta c Bi ton 3.Bi ton 3: Chng minh rng trong tam gic ABC nhn ta lun c:

1 1 1 1 1 1 1

2a b c

l l l R A B C

+ + > + +

Mt khc, ta li c( )2 sin sin

A2 os 2sin

22 2

a

R B Cbc b c

Alc

++= =

. p dng bi ton i s ta

c:

( )( )2

2 2a

B CRR B C bc

AA l

+

+> >

( ) ( )

( )

4

a

R B C R B Cbc

B C l B C

+ +> >

+ +

4

a

bc RR

l

> > .

Hon ton tng tta c:4

c

ab RR

l

> > v

4

b

ca RR

l

> > . Ty, cng 3 chui bt

ng thc ta c:Bi ton 4: Chng minh rngtrong tam gic ABC nhn ta lun c:

12 3c a b

R ab bc ca Rl l l

< + + <

Trong tam gic ta c kt qu sin b ch h

Ac b

= = , sin c ah h

Ba c

= = v sin a bh h

Cb a

= = ,

m t kt qu ca bi ton i s ta d dng c 2 sin sin sinA B C < + + < , m

( )1 1

2 sin sin sin aA B C hb c

+ + = +

1 1 1 1b ch h

c a a b

+ + + +

, t y ta c c Bi

ton 5.Bi ton 5: Chng minh rngtrong tam gic ABC nhn ta lun c:

1 1 1 1 1 14 2a b ch h h

b c c a a b

< + + + + + + + +

( )4 2R r aA bB cC > + +

Kt hp 2 iu trn ta c iu phi chng minh.Sau y l cc bi ton c hnh thnh tcc cng thc quen thuc cc bn luyn

tp:Bi ton: Chng minh rng trong tam gic ABC nhn ta lun c:a/ ( ) ( )2 8 2 2p R r aA bB cC p R r + < + + < + .

b/ ( ) ( ) ( )( ) ( ) ( ) 22

Sp a p b p b p c p c p a S

< + + < .

c/ ( ) ( ) ( )2 2 22

abc a p a b p b c p c abc

< + + < .

d/1 1 1 1 1 1

4 2a b cl l lb c c a a b

< + + + + + = ( )' 0f x > nn hm ( )f x ng bin .

Ch 3 bt ng thc i s:1.Bt ng thc AM-GM:

Cho nsthc dng 1 2, , ..., na a a , ta lun c:1 2

1 2

......n n

n

a a aa a a

n

+ + +

Du = xy ra 1 2 ... na a a = = = .

2.Bt ng thc Cauchy-Schwarz:

Cho 2 bns ( )1 2, ,..., na a a v ( )1 2, ,..., nb b b trong 0, 1,ib i n> = . Ta lun c:


86/106




( )222 2

1 21 2

1 2 1 2

......

...nn

n n

a a aaa a

b b b b b b

+ + ++ + +

+ + +

Du = xy ra 1 2

1 2

... n

n

aa a

b b b = = = .

3.Bt ng thc Chebyshev:Cho 2 dy ( )1 2, ,..., na a a v ( )1 2, ,..., nb b b cng tng hoc cng gim, tc l:

1 2

1 2

...

...n

n

a a a

b b b

hoc 1 2

1 2

...

...n

n

a a a

b b b

, th ta c:

1 1 2 2 1 2 1 2... ... ....n n n na b a b a b a a a b b b

n n n

+ + + + + + + + +

Du = xy ra 1 2

1 2

...

...n

n

a a a

b b b

= = =

= = =.

Nu 2 dy n iu ngc chiu th i chiu du bt ng thc.Xt trong tam gic BC c A B (A,B s o hai gc A,B ca tam gic theo

radian).

A B sin sin

A B

A B ( theo chng minh trn th hm ( )

xf x =

sinx)

2 2

A B

a b

R R

A a

B b , m A B a b . Nhvy ta suy ra nu a b th

A a

B b

(i).

Hon ton tng t : a b c

A B C

a b c v nh vy ta c

( )A

0B

a ba b

, ( ) 0

B Cb c

b c

v ( ) 0

C Ac a

c a

.Cng 3

bt ng thc ta c ( ) 0cyc

A Ba b

a b

( ) ( )2

cyc

AA B C b c

a+ + + (1).

-Cng A B C+ + vo 2 v ca (1) ta thu c:

( ) ( )3A B C

A B C a b ca b c

+ + + + + +

(2)

-TrA B C+ + vo 2 vca (1) ta thu c: ( ) ( )2cyc

AA B C p a

a

+ + (3).

Ch rng A B C + + = v 2a b c p+ + = nn (2) 3 2cyc

Ap

a

3

2cyc

A

a p

(ii), v (3) ( )

2cyc

Ap a

a

(iii).


87/106




Mt khc ta c thp dng bt ng thc Chebyshev cho 2 bs

, ,A B C

a b c

v ( ), , .p a p b p c Ta c: a b c A B C

a b c

p a p b p c

( )( )

3 3 3cyc

A A B Cp ap a p b p ca a b c

+ + + +

( )3

cyc

cyc

ApaA

p aa

. M

3

2cyc

A

a p

ta suy ra: ( )

32

3 3cyc

cyc

Ap p

aA pp a

a

hay ( )

3 2cyc

cyc

Ap

aAp a

a

(iv).

Ta ch n hai bt ng thc (ii) v (iii):

-p dng bt ng thc AM-GM cho 3 s , ,A B C

a b cta c:

13. .

3. .cyc

A A B C

a a b c

kt

hp vi bt ng thc (ii) ta suy ra13. . 3

3. . 2

A B C

a b c p

3. . 2

. .

a b c p

A B C

(v). Mt

khc, ta li c

1

3. .3

. .cyc

a a b c

A A B C

, m theo (v) ta ddng suy ra

1

3. . 2

. .

abc p

ABC

, t ta

c bt ng thc6

cyc

a p

A (vi).

-p dng bt ng thc Cauchy-Schwarz , ta c :

( )22 2

cyc cyc

A B CA A

a aA Aa Bb Cc Aa Bb Cc

+ += =

+ + + + (vii), m ta tm c

( ) ( )2 8 2 2p R r Aa Bb Cc p R r + < + + < + (bi tp a/ phn trc) nn

( )

2

2cyc

A

a p R r

>

(viii) (chng vi tam gic nhn).

-p dng bt ng thc AM-GM cho 3 s ( ) ( ) ( ), ,A B C

p a p b p ca b c

ta c:

( ) ( ) ( ) ( ) ( ) ( )2

3 3 3. . . . . . .

3 3 3. . 4 . 4 .

A B C ABC S ABC S ABCp a p b p c p a p b p c

a b c abc p S R p R + + = =

( )2

3. .

34 .cyc

A S A B Cp a

a p S R (4)m ( )

3 2cyc

cyc

Ap aA

p aa

(theo iv) nn t (4)

32

43

. . 729 . . .3

4 . 3 2 4cyc

cyc

Ap

aS A B C S A B C Ap

p S R R a

3

4729 . . . 3

4 2

S A B C p

R p

354 . . . . .S A B C p R (ix).


88/106




Xt tng

2 22y yx z x z

Tb By a Ax a Ax c Cz c Cz b By

= + + + + +

.

Ta c: 0T

2 2 2

1 1 1 1 1 1. . . 2 0y z z x x y

x a A y b B z c C ab AB bc BC ca CA

+ + + + + + +

.

. . . 2 0y z bc z x ca x y ab c a b

x aA y bB z cC AB BC CA

+ + + + + + +

. . . 2y z bc z x ca x y ab a b c

x aA y bB z cC BC CA AB

+ + + + + + +

(5).

p dng bt ng thc AM-GM ta c:

13 6

3a b c abc p

ABCBC CA AB

+ +

(6).

T(5) v (6) ta c: 6. . .y z bc z x ca x y ab px aA y bB z cC + + ++ + (7).

Thay (x, y, z) trong (7) bng (p-a, p-b, p-c) ta c:

( ) ( ) ( )

12bc ca ab p

A p a B p b C p c + +

(x)

Thay (x, y, z) trong (7) bng (bc, ca, ab)ta c:12b c c a a b p

A B C

+ + ++ + (xi).

3/ Chng ta xt bt ng thc sau:2x

sinx

vi

x

0, 0, 0, 0,2222

(phn chng minh bt

ng thc ny dnh cho bn c).

Theo nh l hm ssin ta c sin2

aA

R= v kt hp vi bt ng thc trn ta c

2 4

2

a A a R

R A , t ta ddng suy ra

12

cyc

a R

A > .

4/ Bt ng thc:2 2

2 2

sin x - x

x + x vi ( ]x 0, 0, 0, 0, (bt ng thc ny xem nhbi

tp dnh cho bn c).

Bt ng thc trn tng ng2

2 2

sin 21

x x

x x

+

3

2 2

2sin

xx x

x

+(1).

Trong tam gic ta c: 3 3sin sin sin2

A B C+ + (2) (bn c tchng minh).T(1)

v (2) ta thu c3 3 3

2 2 2 2 2 2

3 3sin 2

2 cyc

A B CA A B C

A B C

> + + + +

+ + +

3 3 3

2 2 2 2 2 2

3 32

2

A B C

A B C

> + +

+ + +

3 3 3

2 2 2 2 2 2

3 3

2 4

A B C

A B C

+ + >

+ + +.


89/106




Mt khc, p dng bt ng thc cho 3 gc A, B, C ta thu c2 2

2 2

sinA A

A A

>

+,

2 2

2 2

sinB B

B B

>

+ v

2 2

2 2

sin C C

C C

>

+, cng cc bt ng thc ta c:

2 2 2 2 2 22 2 2 2 2 2

sin sin sinA B C A B CA B C A B C

+ + > + ++ + +

, t y p dng nh l hm s sin

sin2

aA

R= ta c

2 2 2 2 2 2

2 2 2 2 2 22 2 2

a b c

A B CR R R

A B C A B C

+ + > + +

+ + +hay

2 2

2 22

cyc

a AR

A A

>

+ .


90/106




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