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7/30/2019 Bt ng thc lng gic

1/101

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

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Chuyn :

BBBTTT NNNGGG TTTHHHCCC LLLNNNGGG GGGIIICCCTHPT Chuyn L T Trng. Cn Th


2/101

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


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 )


3/101


4/101



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 :


5/101



( ) ( )( )( )

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 :


6/101



( )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 :


7/101



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


8/101



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.


9/101



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.


10/101



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.


11/101



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)(''


12/101



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 :


13/101



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


14/101



++++

+=++

CBACBA

CBACBA

222

222

sinsinsinsinsinsin

coscoscos22sinsinsin

v2

33sinsinsin ++ CBA

2

33sinsinsin2 ++


15/101



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


16/101



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


17/101



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


18/101



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

=++

+=+=++

+=++

=++

==++


19/101



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

+


20/101



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


21/101



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

( ) ( ) ( ) ( )

abcabab

cabafbfbac

=

cossinsin

cos:;:

pcm.

V d 1.3.1.2.

Vi ba


22/101



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


23/101



( ) 111; +


24/101



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

+++=

+++++


25/101



( ) 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


26/101



( )

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

=


27/101



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 :

( )

( )


28/101



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


29/101

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, nn dng phngphp no chng minh. Lc vic chng minh bt ng thcmi thnh cng c.

Nh vy, c th ng u vi cc bt ng thc lng gic,bn c cn nm vngcc phngphp chng minh. s l kim 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 diu ca hm s 572.6. Bi tp . 64


30/101



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 sbin i qua li gia cc btng thc. c ths dng

tt phngphp ny bn c cn trang b cho mnh nhng kin thc cn thit vbin 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 btng thc cn chng minh vdng btng thc ng hay quen thuc. Ngoi ra, ta cng c ths dng hai ktququen 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


31/101



V zyx ,, i mt khc nhau nn ( )4 ng pcm.

Nhvy, vi cc btng thc nh trn th vic bin i lng gic l quytnhsng cn vi vic chng minh btng thc. Sau khi s dng cc bin i th vicgiiquyt btng thc trnn d dng thm ch l hin nhin (!).

V d 2.1.2.

CMR : ( )xbcxcaxabcba sin2cos3sin2222 +++

Li gii :

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

( )( )

( ) ( ) 0cos2sinsin2cos

0coscos2sin22sin

sin22cos2sin2cos2sin2cossin22cos2

cos2sin2cossin2cossin2cos2sin

22

2222

22222

2222222

+

++

+++

+

++++++

xbxacxbxa

xbxxabxa

xbcxcaxxabcxbxaxbcxca

xxxxabcxxbxxa

Bt ng thc cui cng lun ng nn ta c pcm.

V d 2.1.3.

CMR vi ABC btk ta c :

4

9sinsinsin 222 ++ CBA

Li gii :


( )

( )

( )( ) 0sin

4

1

2

coscos

04

1coscoscos

04

12cos2cos

2

1cos

4

9

2

2cos1

2

2cos1cos1

2

2

2

2

2

+

+

+++

+

+

CBCB

A

CBAA

CBA

CBA

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


32/101



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 btk 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,,



33/101



( )( )

( ) ( )( ) ( ) ( ) 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


34/101



do

zyx Khi theo AM GM th :

( )( )( ) ( )( )( )( )( )( )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 skt hp hon gia i s v hnh hc. Nhng tnh chtca vectorli mang

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

V d 2.3.1.


45/101



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.


46/101



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 :

Vni dung cng nhcch thc s dng cc btng thc chng ta bn chng1: Cc bc u cs. V th phn ny, taskhng 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==++


47/101



( )

2sin

2sin

2sin

2cos2sin2cos2sin2cos2sin

2

3

2sin

2sin

2sin2

2cos

2sin

2cos

2sin

2cos

2sin

2sin

2sin

2sin

sinsinsin4

1

3

CBA

CCBBAA

CBA

CCBBAA

CBA

CBA

++

=

++

=

Suy ra :

( )12

cot2

cot2

cot2

92

sin

2

sin

2

sin

2cos

2sin

2cos

2sin

2cos

2sin

2sin

2sin

2sin

2

9

2cot

2cot

2cot

2sin

2sin

2sin

3

3

CBA

CBA

CCBBAACBA

CBACBA

=

++

++

m ta cng c : 332

cot2

cot2

cot CBA

( )22

3933

2

9

2cot

2cot

2cot

2

9 33 =CBA

T ( )1 v ( )2 :

2

39

2

cot

2

cot

2

cot

2

sin

2

sin

2

sin

++

++

CBACBA

pcm.

V d 2.4.2.

Cho ABC nhn. CMR :

( )( )2

39tantantancoscoscos ++++ CBACBA

Li gii :V ABC nhn nn CBACBA tan,tan,tan,cos,cos,cos u dng.

Theo AM GM ta c : 3 coscoscos3

coscoscosCBA

CBA

++

CBA

CBACBACBA

coscoscos

sinsinsintantantantantantan ==++


48/101



( )

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.


49/101



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 :


50/101



( )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 :


51/101



( ) ( )( )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 btk. 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

+

+

+


52/101



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

+

=

+

=

+

=


53/101



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

+

+


54/101



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


55/101



+

++

+

+

++

+

+=++

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, khosthm sca chng trnh 12 THPT. Phngphp ny thc s c hiu qu trong cc bi btngthc lnggic. c ths 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 :


56/101



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 :


57/101



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

2222222+

=+

Nh vy 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.


58/101



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 nh nhau nn ta c th gi s cba

Theo gi thit :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>


59/101



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 d 2.5.7.


60/101



CMR vi mi ABC ta c :

( )( )( ) 233

38222 eRcRbRaR


61/101



( ) ( ) ( )[ ]

( ) ( ) 1coscoscoscos69

2cos2cos22

1coscos69

22

22

+++=

++++=

BACBAC

BABABAC

do ( ) 1cos BA

( )22

cos3coscos69 CCCP =+ m 0cos >C

( ) ( ) ( )CCCP 222 cos1cos3cos1 ++

Mt khc ta c :2

1cos600 0 < CC

Xt ( ) ( ) ( )22 13 xxxf += vi

1;

2

1x

( ) ( )( )( )

= 1;

2

1012132' xxxxxf

( )xf ng bin trn khong .( ) ( )( )( ) +++=

16

125cos1cos1cos1

16

125

2

1 222 CBAfxf pcm.

V d 2.5.9.

Cho ABC bt k. CMR :

( ) 32cotcot

sin

1

sin

12 +

+ CB

CB

Li gii :

Xt ( ) xx

xf cotsin

2= vi ( );0x

( ) ( )3

0'sin

cos21

sin

1

sin

cos2'

222

==

=+= xxf

x

x

xx

xxf

( ) 3cotsin

23

3max =

= x

xfxf

Thay x bi CB, trong bt ng thc trn ta c :

3cotsin

2

3cotsin

2

CC

BB pcm.


62/101



V d 2.5.10.

CMR :20

720sin

3

1 0


63/101



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


77/101

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

lin quan n btng 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.Mrng 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 ++++


78/101




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 givi 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


79/101




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


80/101




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 s xc 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 t chng 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, S ln 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

< + + <


81/101




Nhn xt:

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

A B BR

+ + = v

bi ton i s ta d dng 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 nh sau.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 t nh nhn 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 t cng 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= = +


82/101




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

+ + > + +

.

Nh vy 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 t ta 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 t cc 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 n s thc 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 bn s ( )1 2, ,..., na a a v ( )1 2, ,..., nb b b trong 0, 1,ib i n> = . Ta lun c:


85/101




( )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 . Nh vy 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:( ) ( )3

A B CA B C a b c

a b c

+ + + + + +

(2)

-TrA B C+ + vo 2 v ca (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).


86/101




Mt khc ta c th p dng bt ng thc Chebyshev cho 2 b s

, ,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 d dng 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).


87/101




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 s sin 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 d dng 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 t chng 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

+ + >

+ + +.


88/101




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

>

+ .


89/101




Thtrv ci ngun ca mn lng gicL Quc Hn

i hc Sphm Vinh

Lng gic hc c ngun gc t Hnh hc. Tuy nhin phn ln hc sinh khi hcmn Lng gic hc (gii phng trnh lng gic, hm s lng gic ), li thy nnh l mt b phn ca mn i s hc, hoc nh mt cng c gii cc bi ton hnhhc (phn tam gic lng) m khng thy mi lin h hai chiu gia cc b mn y.

Trong bi vit ny, ti hy vng phn no c th cho cc bn mt cch nhn mi :dng hnh hc gii cc bi ton lng gic.

Trc ht, ta ly mt kt qu quen thuc trong hnh hc scp : Nu G l trng tmtam gic ABC v M l mt im ty trong mt phng cha tam gic th :

( ) ( )22222229

1

3

1cbaMCMBMAMG ++++= (nh l Lp-nt)

Nu OM l tm ng trn ngoi tip ABC th 2222 3RMBMBMA =++ nn p

dng nh l hm s sin, ta suy ra : ( )CBARROG 222222 sinsinsin9

4++=

( ) ( )1sinsinsin4

9

9

4 22222

++= CBAROG

Tng thc ( )1 , suy ra :

( )24

9sinsinsin 222 ++ CBA

Du ng thc xy ra khi v ch khi OG , tc l khi v ch khi ABC u.Nh vy, vi mt kin thc hnh hc lp 10 ta pht hin v chng minh c bt ng

thc ( )2 . Ngoi ra, h thc ( )1 cn cho ta mt ngun gc hnh hc ca bt ng thc( )2 , iu m t ngi nghn. Bng cch tng t, ta hy tnh khong cch gia O vtrc tm H ca ABC . Xt trng hp ABC c 3 gc nhn. Gi E l giao im caAH vi ng trn ngoi tip ABC . Th th :

( ) HAHEROHOH .22

/ ==

Do : ( )*.22 HEAHROH = vi :

ARC

ACR

C

AAB

C

AFAH cos2

sin

cossin2

sin

cos.

sin====

v CBABCBKHKHE cotcos2cot22 ===

CBRC

CBCR coscos4

sincoscossin2.2 ==

Thay vo ( )* ta c :

( )3coscoscos8

18 22

= CBAROH


90/101




Nu 090=BAC chng hn, th ( )3 l hin nhin. Gi s ABC c gc A t. Khi

( ) HEHAOHROH .22

/ == trong ARAH cos2= nn ta cng suy ra ( )3 .

T cng thc ( )3 , ta suy ra :

( )48

1

coscoscos CBA (Du ng thc xy ra khi v ch khi ABC u). Cngnh bt ng thc ( )2 , bt ng thc ( )4 c phthin v chng minh ch vi kin thc lp 10 v c mtngun gc hnh hc kh p. Cn nh rng, xanay cha ni n vic pht hin, ch ring vic chngminh cc bt ng thc , ngi ta thng phi dngcc cng thc lng gic (chng trnh lng gic lp11) v nh l v du tam thc bc hai.C c ( )1 v ( )3 , ta tip tc tin ti. Ta th s dng ng thng le.

Nu O, G, H l tm ng trn ngoi tip, trng tm v trc tm ABC th O, G, Hthng hng v : OHOG

3

1= . T 22

9

1OHOG = .

T ( )( )31 ta c :

( ) ( )CBACBA coscoscos814

1sinsinsin

4

9 222 =++

hay CBACBA coscoscos22sinsinsin 222 +=++ Thay 2sin bng 2cos1 vo ng thc cui cng, ta c kt qu quen thuc :

( )51coscoscos2coscoscos 222 =+++ CBACBA

Cha ni n vic pht hin ra ( )5 , ch ring vic chng minh lm nhc c khngbit bao nhiu bn tr mi lm quen vi lng gic. Qua mt vi v d trn y, hn ccbn thy vai tr ca hnh hc trong vic pht hin v chng minh cc h thc thunty lng gic. Mt khc, n cng nu ln cho chng ta mt cu hi : Phi chng cc hthc lng gic trong mt tam gic khi no cng c mt ngun gc hnh hc lm bnng ? Mi cc bn gii vi bi tp sau y cng c nim tin ca mnh.

1. Chng minh rng, trong mt tam gic ta c

=

2sin

2sin

2sin8122

CBARd trong

d l khong cch gia ng trn tm ngoi tip v ni tip tam gic .T hy suy ra bt ng thc quen thuc tng ng. 2. Cho ABC . Dng trong mt phng ABC cc im 1O v 2O sao cho cc tam

gic ABO1 v ACO2 l nhng tam gic cn nh 21,OO vi gc y bng 030 vsao cho 1O v C cng mt na mt phng bAB, 2O v B cng mt na mtphng bAC.

a) Chng minh :

( )ScbaOO 346

1 222221 ++=

b) Suy ra bt ng thc tng ng :


91/101




CBACBA sinsinsin32sinsinsin 222 ++

3. Chng minh rng nu ABC c 3 gc nhn, th :

2coscoscos

sinsinsinBA

v 02

3cos >

+

C

.

Ta c :

( )74

sin22

sin2

sin4

sin2

cos18

1

2cos2cos18

1

2

coscos

18

1

2

2sin

2sin

2

2sin

2sin

66663

3

32266

BABABABA

BABABA

BABA

++

+=

+

+

=

+

=

+

+

Tng t ta c : ( )84

3sin223sin

2sin 666

+

+

CC

Cng theo v ca ( )7 v ( )8 ta c :

( )964

3

6sin3

2sin

2sin

2sin

83sin4

43sin

4sin2

23sin

2sin

2sin

2sin

6666

6666666

=++

+++

++

++++

CBA

CBACBACBA

Trng hp tam gic ABC nhn, cc bt ng thc ( ) ( ) ( )9,8,7 lun ng.Th d 4. Chng minh rng vi mi tam gic ABC ta lun c :

( )( )( )

3

4

6

4

222sincossincossincos

++++ CCBBAA

Li gii. Ta c :

( )( )( )

=+++

4cos

4cos

4cos22sincossincossincos

CBACCBBAA

nn bt ng thc cho tng ng vi :


95/101




( )*4

6

4

2

4cos

4cos

4cos

3

+

CBA

- Nu { }4

3,,max

CBA th v tri ca ( )* khng dng nn bt ng thc cho

lun ng.

- Nu { }4

3,,max

>

>

CBA

nn ( )

+

+=

BABABA cos

2cos

2

1

4cos

4cos

( )1042

cos4

cos4

cos

42cos

2cos1

2

1

2

2

+

+

++

BABA

BABA

Tng t :

( )1142

3cos43

cos4

cos 2

+

C

C

Do nhn theo v ca ( )10 v ( )11 ta s c :

+

+

43cos

423cos

42cos

43cos

4cos

4cos

4cos 422

C

BACBA

3

3

4

6

4

2

43cos

4cos

4cos

4cos

+=

CBA

Do :

( )( )( )

3

4

6

4

222sincossincossincos

++++ CCBBAA

ng thc xy ra khi v ch khi tam gic ABC u.Mi cc bn tip tc gii cc bi ton sau y theo phng php trn.

Chng minh rng vi mi tam gic ABC, ta c :

( )NnCBA

CBA

n

nnn

++

++

2.3

2sin

1

2sin

1

2sin

1)2

3

1

2tan

2tan

2tan)1 333


96/101




( )314

2

4cos

4cos

4cos)3 +++

CC

BB

AA

( ) CBACBA coscoscos3122

1

4cos

4cos

4cos)4

3+

vi ABC nhn.

B SUNG:Cm n cc bn THPT chuyn L T Trng dn bi ny! Tc gi bi nycng chnh l ch nhn ca trang web www.laisac.page.tl , nn mn php xin b sungthm bi vit ny vi mc na cho sinh ng , ch t cng l sinh s he he he

GII CC BI NGH

Bi 1. Chng minh rng vi mi tam gic ABC , ta u c

31

222333 ++ CtgBtgAtg

Li gii . Ta c

333

222

222

+

+

Btg

Atg

Btg

Atg

.

Mt khc:

22

Btg

Atg + = .

42

2cos1

4cos

4sin4

2cos2cos

2sin2

2cos2cos

2sin

BAtg

BA

BABA

BABA

BA

BA

BA+

=+

+

++

+

+

-

+

=

+

Do :4

222

333 BAtgB

tgA

tg+

+ (14). ( C dng

++

22)()(

BAfBfAf )

Tng t4

602

2

60

2

03

033 ++

Ctgtg

Ctg (15)

Cng theo v (14) v (15) ta c:

2

604)

4

60

4(2

2

60

222

03

033

03333 tg

Ctg

BAtgtg

Ctg

Btg

Atg

++

+=+++

.

3

1303

222

03333 =++ tgC

tgB

tgA

tg

ra khi v ch khi tam iaca ABC u.ng thc xy


97/101

Bi 2. Chng minh rng vi mi tam gic ABC , ta u cn

nnn CBA2.3

2sin

1

2sin

1

2sin

1++ ( n l s thc dng)

Li gii . Ta c:

4sin

2

)2

cos1(

2

)2

cos2

(cos

2

)2

sin2

(sin

1.2

2sin

1

2sin

1

2

2

2

2

2

2

2BABABABABABA nn

n

n

n

nnn +

=+

-

+

--

=

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