Chuyên đề bất đẳng thức bồi dưỡng học sinh giỏi Toán 8 - THCS.TOANMATH.com

1

GV: Ngô Thế Hoàng _ THCS Hợp Đức

CHUYÊN ĐỀ BẤT ĐẲNG THỨC

DẠNG 1: SỬ DỤNG ĐỊNH NGHĨA: A>B TA XÉT HIỆU A-B >0, CHÚ Ý BĐT A² 0 Bài 1: CMR : với mọi x,y,z thì x²+y²+ z² xy+ +yz zx

HD:

Xét hiệu ta có:

(

^{2 2 2}

) ( ) (

²

) (

²

)

²

2 x +y + −z xy−yz−zx  =0 x−y + y−z + −z x 0 Dấu bằng xảy ra khi x = y = z

Bài 2: CMR : với mọi x,y,z thì x²+y²+ z² 2xy+2yz−2zx HD:

Xét hiệu ta có:

( )

²

2 2 2

2 2 2 0 0

x +y +z − xy− yz+ zx = x− +y z  Dấu bằng xảy ra khi x+z=y

Bài 3: CMR : với mọi x,y,z thì ^{x2+y2+ + z2 3 2}

(

^{x+ +y z}

)

HD:

Xét hiệu ta có:

(

^x−1

) (

^{2+ y−1}

) (

^{2+ −z 1}

)

^{2 0}

Dấu bằng khi x=y=z=1 Bài 4: CMR : với mọi a,b ta có :

2 2 2

2 2

a +b  a b+  HD :

Xét hiệu ta có :

2 2 2 2

2 0

2 4

a +b −a + ab b+  <=>^2a2+2b2−

(

^{a2−2ab b+ 2}

)

^0

( )

²

2 2

2 0 0

a ab b a b

= + +  = +  Dấu bằng khi a=b

Bài 5: CMR : với mọi a,b,c ta có :

2 2 2 2

3 3

a +b +c a b c+ + 

   HD:

Ta có:

2 2 2 2 2 2

2 2 2

3 9

a + +b c a + + +b c ab+ bc+ ac

( )

2 2 2 2 2 2

3a 3b 3c a b c 2ab 2bc 2ac 0

= + + − + + + + + 

2 2 2

2a 2b 2c 2ab 2bc 2ac 0

= + + − − − 

(

^{a b}

) (

^{2 b c}

) (

^{2 c a}

)

^{2 0}

= − + − + −  , Dấu bằng khi a=b=c

Bài 6: CMR : 2 2 2

( )

²

3 a b c

a b c + +

+ +  HD:

Ta có:

2 2 2 2 2 2

3a +3b +3c a +b + +c 2ab+2bc+2ca

2 2 2

2a 2b 2c 2ab 2bc 2ac 0

= + + − − − 

(

^{a b}

) (

^{2 b c}

) (

^{2 c a}

)

^{2 0}

= − + − + −  , Dấu bằng khi a=b=c

2

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 7: CMR : 2 2

( )

²

2 2

a b a b+ ab

+  

HD:

Ta chứng minh: 2 2

( )

²

2 a b a b+

+ 

2 2 2 2

2a 2b a 2ab b

= +  + +

( )

²

2 2

2 0 0

a b ab a b

= + −  = −  Dấu bằng khi a=b

Ta chứng minh

( )

²

2 2

a b+ ab



( )

²

2 2

2 4 0

a ab b ab a b

= + +  = −  Dấu bằng khi a=b

Bài 8: Cho a,b,c là các số thực, CMR:

2 2

4

a +b ab HD:

Ta có:

( )

²

2 2

4a +b −4ab= 2a b− 0 Dấu bằng khi b=2a

Bài 9: Cho a,b,c là các số thực, CMR : a²+b²+ 1 ab a b+ + HD:

Ta có:

2 2

1 0

a +b + −ab a b− − 

2 2

2a 2b 2 2ab 2a 2b 0

= + + − − − 

(

^{a2 2ab b2}

) (

^{a2 2a 1}

) (

^{b2 2b 1}

)

⁰

= − + + − + + − + 

(

^{a b}

) (

^{2 a 1}

) (

^{2 b 1}

)

^{2 0}

= − + − + −  Dấu bằng khi a=b=1

Bài 10: Cho a,b,c,d là các số thực : CMR : ^{a2+ + +b2 c2 d2+ e2 a b c d}

(

^{+ + +e}

)

HD:

Ta có:

2 2 2 2 2

0 a +b + +c d + −e ab ac ad− − −ae

2 2 2 2 2

4a 4b 4c 4d 4e 4ab 4ac 4ad 4ae 0

= + + + + − − − − 

=

(

^a2−4ab+4b2

) (

^{+ a2−4ac+4c2}

) (

^{+ a2−4ad+4d2}

) (

^{+ a2−4ae+4e2}

)

^0

=

(

^a−2b

) (

^{2+ a−2c}

) (

^{2+ a−2d}

) (

^{2+ a−2e}

)

^20

Dấu bằng xảy ra khi a=2b=2c=2d=2e

Bài 11: Cho a,b thỏa mãn: a+b = 1, a>0, b>0 CMR: 1 1

1 1 9

a b

 +  + 

  

   HD:

ta có: VT 1 a b 1 a b 2 b 2 a 4 2 a b 1

a b a b b a

+ +

       

= +  +   = +  + = +  + +

5 2 a b 5 2.2 9

b a

 

= +  +  + =

 

Dấu bằng khi ^{2 2 1}

2 a b

a b a b

b = =a + = = =

3

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 12: Cho

2

, 0, :

2 x y

x y CMR  +  xy HD:

Ta có:

( )

²

2 2 2 2

2 4 2 0 0

x +y + xy xy=x − xy+y  = x−y  , Dấu bằng khi x=y Bài 13: Cho a > 0, b > 0, CMR: a³+b³ a b ab² + ²

HD:

Ta có:

(

^{a3−a b2}

) (

^{+ b3−ab2}

)

^{ =0 a2}

(

^{a b− −}

)

^b2

(

^{a b−}

)

^0

=

(

^{a b a−}

) (

^2−b2

)

^{ =0}

(

^{a b−}

) (

^{2 a b+ }

)

⁰

Dấu bằng khi a=b

Bài 14: Cho a b 1, CMR: ¹ ₂ ¹ ₂ ² 1 a +1 b 1 ab

+ + +

HD:

Xét hiệu:

2 2

1 1 1 1

1 a 1 ab 1 b 1 ab 0

 −  + − 

 + +   + + 

   

( )

(

^{1 a b aa2}

) (

^{−1 ab}

) (

^{1 b a bb}

(

²

) (

⁻¹

)

^ab

)

⁰

= + 

+ + + +

=

( ) ( )

( ) ( )( )

2

2 2

1 0

1 1

b a ab

ab a b a

− −

+ + + 

Dấu bằng khi a=b hoặc a=b=1

Bài 15: CMR : với mọi số thực x,y,z,t ta luôn có : ^x2+^y2+ + ^{z2 t2 x y}

(

+ +^{z t}

)

HD:

Ta có:

2 2 2 2

0 x +y + + − − − z t xy xz xt

= 4x²+4y²+4z²+4t²−4xy−4xz−4xt0

=

(

^x2−4xy+4y2

) (

^{+ x2−4xz+4z2}

) (

^{+ x2−4xt+4t2}

)

^{+x2 0}

Dấu bằng khi x= 2y=2z=2t=0 Bài 17: CMR :

2

2 2

4 2

a + + b c ab ac− + bc HD:

Ta có:

2 2 2

4 4 4 4 8 0

a + b + c − ab+ ac− bc

( ) ( )

2 2 2

4 4 2 0

a a b c b c bc

= − − + + − 

= ^{a2−4a b c}

(

^{− +}

) (

^{4 b c−}

)

^{2 0}

=

(

^a−2a+2c

)

^20

Bài 19: CMR : x²+y²+ z² 2xy−2zx+2yz HD:

Ta có:

2 2 2

2 2 2 0

x +y + −z xy− yz+ zx

( )

2 2 2

2 2 0

x − x y− +z y − yz+z 

( ) ( )

²

( )

²

2 2 0 0

x − x y− +z y−z  = x− +y z 

4

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 20: CMR : ^{x4+y4+z4+ 1 2x xy}

(

^{2− − +x z 1}

)

HD:

Ta có:

4 4 4 2 2 2

1 2 2 2 2 0

x +y + + −z x y + x − xz− x

(

^{x4+y4−2x y2 2}

) (

^{+ x2−2xz+z2}

) (

^{+ x2−2x+ 1}

)

⁰

(

^x2−y2

)

^{2+ −}

(

^{x z}

) (

^{2+ −x 1}

)

^{2 0}

Dấu bằng khi x=z=1, y=1 Bài 21: CMR : a²+b²+c² ab bc+ +ca HD:

Ta có :

2 2 2

0 a +b + −c ab bc ca− − 

2 2 2

2a 2b 2c 2ab 2bc 2ca 0

= + + − − − 

(

^{a b}

) (

^{2 b c}

) (

^{2 c a}

)

^{2 0}

= − + − + −  Bài 22: CMR : a²+b²ab

HD:

ta có:

2 2

0 a +b −ab

2 2 2 2

2 3 3

2 . 0 0

2 4 4 2 4

b b b b b

a a a 

= − + +  = −  + 

 

Bài 23: CMR : x²+ +xy y²0 HD:

Ta có:

2 2 2 2

2 3 3

2 . 0 0

2 4 4 2 4

y y y y y

x + x + +  =x+  +  Bài 24: CMR : a a b a c a b c

(

+

)(

+

)(

+ + +

)

b c^{2 2}⁰ HD: =a a b c a b a c

(

+ +

)(

+

)(

+ +

)

b c^{2 2}⁰

=

(

^{a2+ab ac a+}

)(

^{2+ab ac bc+ +}

)

^{+b c2 20}

Đặt

a2 ab ac x bc y

 + + =

 =



Khi đó ta có: ^{x x}

(

^+y

)

^{+y2 =0 x2+xy+y20}

Bài 25: CMR :

(

^a2+b2

)(

^a4+b4

) (

^{ a3+b3}

)

²

HD:

Ta có:

6 2 4 4 2 6 6 3 3 6

2

a +a b +a b +b a + a b +b

=

(

^{a b4 2−a b3 3}

) (

^{+ a b2 4−a b3 3}

)

^0

= ^{a b a b3 2}

(

^{− +}

)

^{a b b a2 3}

(

⁻

)

^0

=

(

^{a b a b−}

) (

^{3 2−a b2 3}

)

^{ =0 a b2 2}

(

^{a b−}

)

^20

Bài 26: CMR :

(

^{a b a+}

) (

^3+b3

) (

^{2 a4+b4}

)

HD:

Ta có:

4 3 3 4 4 4

2 2

a +ab +a b b+  a + b =a⁴−ab³+b⁴−a b³ 0

= a a b³

(

− +

)

b b a³

(

−

)

0 =

(

^a3−b3

) (

^{a b−  =}

)

⁰

(

^{a b−}

)

²

(

^{a2+ab b+ 2}

)

^0

5

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 27: Cho a,b > 0, CMR : ²

(

^a3+b3

)

^

(

^{a b a+}

) (

^2+b2

)

HD:

Ta có:

3 3 3 2 2 3

2a +2b a +ab +a b b+

= a³−a b b² + −³ ab² 0

= ^a2

(

^{a b− +}

)

^{b b a2}

(

^{− }

)

⁰

=

(

^{a b−}

) (

^{2 a b+}

)

^0

Bài 28: Cho a, b > 0, CMR: ⁴

(

^a3+b3

)

^

(

^{a b+}

)

³

HD:

Ta có:

3 3 3 2 2 3

4a +4b a +3a b+3ab +b

=3a³−3a b² +3b³−3ab² 0

=^3a2

(

^{a b− +}

)

^3b2

(

^{b a−}

)

^{ =0 3}

(

^{a b a−}

) (

^2−b2

)

^0

=³

(

^{a b−}

) (

^{2 a b+}

)

^0

Bài 29: Cho a,b,c > 0, CMR: ^a3+ +^{b3 abc}^{ab a b c}

(

+ +

)

HD:

Ta có:

3 3 2 2

a + +b abca b+ab +abc

= a³−a b b² + −³ ab² 0

= ^a2

(

^{a b− +}

)

^{b b a2}

(

^{− }

)

⁰

=

(

^{a b−}

) (

^{2 a b+}

)

^0

Bài 30: CMR:

(

^a2+b2

)

^{2 ab a b}

(

⁺

)

²

HD:

Ta có:

( )

4 2 2 4 2 2 3 2 2 3

2 2 2

a + a b +b ab a + ab b+ =a b+ a b +ab

=

(

^{a4−a b3}

) (

^{+ b4−ab3}

)

^0

= a a b³

(

− +

)

b b a³

(

−

)

0

=

(

^a3−b3

) (

^{a b−  =}

)

⁰

(

^{a b−}

)

²

(

^{a2+ab b+ 2}

)

^0

Bài 31: CMR: ^{a2+ + b2 c2 a b c}

(

⁺

)

HD:

ta có:

2 2 2

0 a +b + −c ab ac− 

= 4a²+4b²+4c²−4ab−4ac0

=

(

^a2−4ab+4b2

) (

^{+ a2−4ac+4c2}

)

^{+2a2 0}

=

(

^a−2b

) (

^{2+ a−2c}

)

^2+2a20

Bài 32: CMR: a²+ + +b² c² d²a b c d

(

+ +

)

HD:

2 2 2 2

0 a +b + +c d −ab ac ad− − 

= 4a²+4b²+4c²+4d²−4ab−4ac−4ad 0

=

(

^a2−4ab+4b2

) (

^{+ a2−4ac+4c2}

) (

^{+ a2−4ad+4d2}

)

^{+a2 0}

=

(

^a−2b

) (

^{2+ a−2c}

) (

^{2+ a−2d}

)

^{2+a2 0}

6

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 33: CMR: ^{2 2 2 3}

( )

a +b + + c 4 a b c+ + HD:

Ta có:

(

²

) (

²

) (

²

)

^{3 0}

a −a + b − +b c − + c 4

= ² 1 ² 1 ² 1

4 4 4 0

a a b b c c

 − +  + − +  + − + 

     

     

=

2 2 2

1 1 1

2 2 2 0

a b c

 −  + −  + −  

     

     

Bài 34: CMR: a⁴+b⁴+ 2 4ab HD:

ta có:

4 4

4 2 0

a +b − ab+ 

= a⁴+b⁴−2a b^{2 2}+2a b^{2 2}−4ab+ 2 0

=

(

^a2−b2

) (

^{2+2 a b2 2−2ab+ 1}

)

⁰

=

(

^a2−b2

)

²⁺²

(

^ab−1

)

^{2 0}

Bài 35: CMR: x⁴−4x+ 5 0 HD:

ta có:

(

^{x4−4x2+ +4}

) (

^{4x2−4x+ 1}

)

⁰

=

(

^x2−2

)

²⁺

(

^2x−1

)

^{2 0}

Không xảy ra dấu bằng Bài 36: CMR: ^{4 1} 0

x − + x 2 HD:

Ta có:

4 2 1 2 1

4 4 0

x x x x

 − +  + − + 

   

   

=

2 2

2 1 1

2 2 0

x x

 −  + −  

   

   

Bài 37: CMR: x³+4x+ 1 3 (x x² 0) HD:

ta có: x³−3x²+4x+ 1 0

= ^{x x}

(

^{2− +x 4}

)

^{+x2+ 1 0}

= ^{x x}

(

⁻²

)

^{2+x2+ 1 0}, Vì x > 0 Bài 39: CMR:

(

x−1

)(

x−2

)(

x−3

)(

x−  −4

)

1 HD:

(

x−1

)(

x−4

)(

x−2

)(

x− + 3

)

1 0

=

(

^x2−5x+4

)(

^{x2−5x+ + 6}

)

^{1 0}

Đặt x²−5x+ =5 t

Khi đó ta có:

( )( )

^{t−1 t+ + 1 1 0}

=t² 0, Dấu bằng khi t=0

7

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 40: CMR: x⁴+x³+x²+ + x 1 0 HD:

Ta có : ^x3

(

^{x+ + + +1}

) (

^{x 1}

)

^x20

=

(

^x+1

) (

^{x3+ +1}

)

^{x2 0x}

=

(

^x+1

)

²

(

^{x2− + +x 1}

)

^{x2 0 ( ĐPCM)}

Bài 41: CMR : a²+4b²+4c²4ab+8bc−4ac HD:

Ta có:

2 2 2

4 4 4 8 4 0

a + b + c − ab− bc+ ac

= ^a2+

( ) ( )

^{2b 2+ 2c 2−2. .2a b−2.2 .2b c+2. .2a c0}

=

(

^{a b c− +}

)

^{2 0}

Bài 42: CMR : ⁸

(

^{a3+ +b3 c3}

)

^

(

^{a b+}

) (

^{3+ +b c}

) (

^{3+ +c a}

)

³ với a, b, c >0 HD:

Ta có:

3 3 3 3 3 3 2 2 2 2 2 2

8a +8b +8c 2a +2b +2c +3a b+3ab +3b c+3bc +3a c+3ac

= 6a³+6b³+6c³−3a b² −3ab²−3b c² −3bc²−3a c² −3ac² 0

=

(

^{3a3−3a b2}

) (

^{+ 3a3−3a c2}

) (

^{+ 3b3−3b a2}

) (

^{+ 3b3−3b c2}

) (

^{+ 3c3−3bc2}

) (

^{+ 3c3−3ac3}

)

^0

=^3a2

(

^{a b− +}

)

^3a2

(

^{a c− +}

)

^{3b b a2}

(

^{− +}

)

^{3b b c2}

(

^{− +}

)

^{3c c b2}

(

^{− +}

)

^{3c c a2}

(

^{− }

)

⁰

=³

(

^{a b a−}

) (

^2−b2

)

⁺³

(

^{a c−}

) (

^a2−c2

)

⁺³

(

^{b c b−}

) (

^2−c2

)

^0

=³

(

^{a b−}

) (

^{2 a b+ +}

) (

^{3 a c−}

)(

^{a c+ +}

) (

^{3 b c−}

) (

^{2 b c+ }

)

⁰

Bài 43: CMR:

(

^{a b c+ +}

)

^{3a3+ + +b3 c3 24abc} với a,b,c>0 HD:

Ta có:

( )( )( )

3 3 3 3 3 3

3 24

a + + +b c a b b c c a+ + + a + + +b c abc

=3

(

a b b c c a+

)(

+

)(

+

)

24abc

Vì

2 2 2 a b ab b c bc c a ca

 + 

 + 

 + 



, Nhân theo vế ta được ĐPCM

Bài 44: CMR: Với mọi x, y # 0 ta có:

2 2

2 2 4 3

x y x y

y x y x

 

+ +   + 

 

HD:

Ta có:

( )

4 4 2 2 2 2

4 3

x +y + x y  xy x +y

=

(

^x2+y2

)

^{2−xy x}

(

^2+y2

)

^{+2x y2 2−2xy x}

(

^2+y2

)

^0

=

(

^x2+y2

)(

^x2+y2−xy

)

^{+2xy xy}

(

^−x2−y2

)

^0

=

(

^{x2 +y2−xy}

)(

^x2+y2−2xy

)

^0

=

(

^x−y

)

²

(

^x2−xy+y2

)

^0

8

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 45: CMR : Nếu a b+ 1, thì ^{3 3 1} a +b  4 HD:

Ta có:

3 2 3

1 1 3 3

b − =a b  − a+ a −a

=

2

3 3 2 1 1 1

3 3 1 3

2 4 4

a +b  a − a+ = a−  +  Bài 46: Cho a,b,c > 0, CMR : ab bc ca+ + a²+b²+c² HD:

Ta có:

2 2 2

0 a +b + −c ab bc ca− − 

=

(

^{a b−}

) (

^{2+ −b c}

) (

^{2+ −c a}

)

^20

Bài 47: CMR :

2 2

1 0 1 a a a a

+ + 

− + HD:

Ta có:

2 2 1 3

1 0,

4 4

a + + =a a + +a +  a

2 2 1 3

1 0,

4 4

a − + =a a − +a +  a

 

Nên VT > 0

Bài 48: CMR : ^{4a a b a}

(

⁺

)(

⁺¹

)(

^{a b+ + + 1}

)

^{b2 0}

HD:

Ta có:

( )( )( )

²

4a a b+ +1 a+1 a b+ +b 0

= ⁴

(

^a2+ab+a

)(

^{a2+ab+ + +a b}

)

^{b2 0. đặt a2 ab a x}

b y

+ + =

=

= ^{4x x}

(

^+y

)

^{+y2 0}

= 4x²+4xy+y²0

=

(

^2x+y

)

^20, Dấu bằng khi 2 2

(

1

)

2 2 2 2

2 1

x y a ab a b b a a

a

= − = + + = − = = − + + Bài 49: CMR : 2 2

( )

²

2 2 x y

x y + xy

+  

HD:

Ta có:

( ) ( )

( ) ( )

2

2 2 2 2 2 2 2

2

2 2 2

2 2 2 0

2

2 2 4 0

2

x y

x y x y x y xy x y

x y

xy x y xy xy x y

 +

+  = +  + + = − 



 +  = + +  = − 



Bài 50: CMR : ^{1 1 4} a+ b a b

+ , Với a,b > 0 HD:

Ta có:

(

a b

)

4 ab a b

+ 

+ ^=

(

^{a b+}

)

^{2 4ab=}

(

^{a b−}

)

^{2 0}

9

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 51: CMR : ^{a4+b4 ab a}

(

^2+b2

)

HD:

Ta có: a⁴+b⁴−a b ab³ − ³ 0

= a a b³

(

− +

)

b a b³

(

− 

)

0

=

(

^{a b−}

)

²

(

^{a2+ab b+ 2}

)

^0

Bài 52: CMR :

4 4 4

2 2

a +b a b+ 

   HD:

Ta có:

4 4 4 4 2 2 2 2 3 3

8a +8b a +b +4a b +2a b +4a b+4ab

= 7a⁴+7b⁴−4a b^{2 2}−2a b^{2 2}−4a b³ −4ab³ 0

=

(

^{a4+b4+2a b2 2}

) (

^{+ 6a4+6b4}

)

^{−4ab a}

(

^2+b2

)

^{−8a b2 2 0}

=

(

^a2+b2

)

^{2−4ab a}

(

^2+b2

)

^{+4a b2 2+6}

(

^a4+b4

)

^{−12a b2 20}

=

(

^a2+b2−2ab

) (

^{2+6 a4+b4−2a b2 2}

)

^0

=

(

^{a b−}

)

⁴⁺⁶

(

^a2−b2

)

^20

Bài 53: Cho a+b+c=0, CMR : ab bc ca+ + 0 HD:

Ta có: a²+ + +b² c² 2

(

ab bc ca+ +

)

=0

= ²

(

^{ab bc ca+ +}

)

^{= −}

(

^a2+b2+c2

)

^0

Dấu bằng khi a=b=c=0

Bài 54: Cho x,y,z R, CMR :

(

^x−y

) (

^{2+ y−z}

) (

^{2+ −z x}

)

^{2 3}

(

^x2+y2+z2

)

HD:

Ta có:

2 2 2 2 2 2

2x +2y +2z −2xy−2yz−2zx3x +3y +3z

= x²+y²+ +z² 2xy+2yz+2zx0

=

(

^{x+ +y z}

)

^{2 0}

Bài 55: CMR : Với mọi x,y khác 0, ta luôn có :

6 6

4 4

2 2

x y x y

y x +  +

HD:

Ta có: ^{x y2 2}

(

^x4+y4

)

^x8+y8

= x⁸+ −y⁸ x y^{6 2}−x y^{2 6}0

= ^x6

(

^x2−y2

)

^−y6

(

^x2−y2

)

^0

=

(

^x6−y6

)(

^x2−y2

)

^0

=

(

^x2−y2

)(

^{x4+x y2 2+y4}

)(

^x2−y2

)

^0

=

(

^{x2 −y2}

) (

^{2 x4+x y2 2+y4}

)

^0

Bài 56: CMR : ^{2a2+ + b2 c2 2a b c}

(

⁺

)

HD:

Ta có: 2a²+b²+ −c² 2ab−2ac0

=

(

^{a2−2ab b+ 2}

) (

^{+ a2−2ac c+ 2}

)

^0

=

(

^{a b−}

) (

^{2+ a c−}

)

^20

10

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 57: CMR : a⁴+a b ab³ + ³+b⁴ 0 HD:

ta có:

( ) ( )

3 3

0 a a b+ +b a b+ 

=

(

^a3+b3

) (

^{a b+}

)

^0

=

(

^{a b+}

)

²

(

^{a2−ab b+ 2}

)

^0

Bài 58: CMR : a⁴−2a b³ +2a b^{2 2}−2ab³+b⁴ 0 HD:

Ta có:

(

^{a4 −2a ab a b2. + 2 2}

) (

^{+ b4−2ab b. 2+a b2 2}

)

^0

(

^{a2 ab}

) (

^{2 b2 ab}

)

^{2 0}

= − + − 

Bài 59: CMR : ^{a4+b4+ + c2 1 2a ab}

(

^{2− + +a c 1}

)

HD:

Ta có:

4 4 2 2 2 2

1 2 2 2 2 0

a +b + + −c a b + a − ac− a

=

(

^{a4+b4−2a b2 2}

) (

^{+ a2−2ac c+ 2}

) (

^{+ a2−2a+ 1}

)

⁰

=

(

^a2−b2

)

²⁺

(

^{a c−}

) (

^{2+ a−1}

)

^{2 0}

Bài 60: CMR :

(

^{ab bc ca+ +}

)

^{2 3abc a b c}

(

^{+ +}

)

HD:

Ta có:

2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 3 3 3 0

a b +b c +c a + ab c+ abc + a bc− a bc− ab c− abc 

2 2 2 2 2 2 2 2 2

0 a b b c c a ab c abc a bc

= + + − − − 

Đặt

ab x bc y ca z

 =

 =

 =

=> x²+y²+ − − − z² xy yz zx 0

=

(

^x−y

) (

^{2+ y−z}

) (

^{2+ −z x}

)

^20

Bài 61: CMR : ^{y 1 1 1}

(

^{x z}

)

^{1 1}

(

^{x z}

)

x z y x z

 + + +  +  +

   

    , Với 0  x y z

HD:

Ta có:

( ) ( )

²

y x z x z x z 0

xz y xz

+ + + − + 

= ^{y2+ −xz y x}

(

^{+ z}

)

⁰

= y²+ − −xz xy yz0

(

y x

)(

z y

)

0

= − − 

Bài 62: Cho a,b dương có tổng 1, CMR : ^{1 1 4}

1 1 3

a +b 

+ +

HD:

Ta có:

Quy đồng =3

(

a b+ + 2

) (

4 a+1

)(

b+1

)

= ⁴

(

^{ab a b}+ + +  = ¹

)

^{9 1 4ab}=

(

^{a b}+

)

²^4ab

=

(

^{a b−}

)

^{2 0 ( đúng)}

11

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 63: CMR : Với a,b,c > 0 thì

2 2

2 2

a b a b

b +a  +b a HD:

Ta có:

2 2

2 2 2 0

a b a b a b

b a b a b a

   

+ −  +  + + 

   

= a₂² b²₂ 2 a b 2

VT b a b a

   

 + −  + +

= a²₂ 2.a 1 b²₂ 2.b 1 0

b b a a

   

− + + − + 

   

   

Bài 64: CMR : ^a8 _{3 3 3}^{b8 c8 1 1 1,}

(

^{a b c, , 0}

)

a b c a b c

+ +  + + 

HD:

Ta có: ^{a8+ + b8 c8 a b4 4+b c4 4+c a4 4 =}

( ) ( ) ( )

^{a b2 2 2+ b c2 2 2+ c a2 2 2}

2 4 2 2 4 2 4 2 2

VT a b c +b c a +a b c ^{=a b c2 2 2}

(

^a2+b2+c2

)

^{a b c2 2 2}

(

^{ab bc ca+ +}

)

8 8 8 8 8 8

2 2 2 3 3 3

1 1 1

a b c a b c

ab bc ca

a b c a b c a b c

+ + + +

=  + + =  + +

Bài 65: CMR :

(

^a10+b10

)(

^a2+b2

) (

^{ a8+b8}

)(

^a4+b4

)

HD:

Ta có: a¹²+a b^{10 2}+a b^{2 10}+b¹²a¹²+a b^{8 4}+a b^{4 8}+b¹²

(

^{a b10 2 a b8 4}

) (

^{a b2 10 a b4 8}

)

⁰

= − + − 

( ) ( )

8 2 2 2 2 8 2 2

0 a b a b a b b a

= − + − 

( )( )

2 2 2 2 6 6

0 a b a b a b

= − − 

= ^{a b2 2}

(

^a2−b2

) (

^{2 a4+a b2 2+b4}

)

^0

Bài 66: Cho a,b,c dương có abc=1, và a b c ^{1 1 1} a b c

+ +  + + , CMR :

(

^a−1

)(

^b−1

)(

^{c− 1}

)

⁰

HD:

Ta có: a b c+ + ab bc ca+ + ,

Xét

(

^a−¹

)(

^b−¹

)(

^c− =¹

)

^abc−

(

^{ab bc ca}+ +

) (

+ + + −^{a b c}

)

¹

=

(

a b c+ + −

) (

ab bc ca+ +

)

0

Bài 67: Cho a,b>0, thỏa mãn : a³+b³= −a b, CMR : a²+b²+ab1 HD:

Ta có:

( ) ( )

3 3 3 3 2 2

a +b a −b = a b a− +ab b+

(

^{a b}

) (

^{a b a}

) (

^{2 b2 ab}

)

= −  − + + =a²+b²+ab1 Bài 68: CMR : ²

(

^a8+b8

) (

^{ a3+b3}

)(

^a5+b5

)

HD:

Ta có: 2a⁸+2b⁸ a⁸+a b^{3 5}+a b^{5 3}+b⁸

=

(

^{a8−a b5 3}

) (

^{+ b8−a b3 5}

)

^0

= ^a5

(

^a3−b3

) (

^{−b5 a3−b3}

)

^0

=

(

^a5−b5

)(

^a3−b3

)

^0, Giả sử a > b => a³b a³, ⁵b⁵ => ĐPCM Nếu a<b => a³b a³, ⁵b⁵ => ĐPCM

12

GV: Ngô Thế Hoàng _ THCS Hợp Đức

Bài 79: CMR : ³

(

^{a8+ +b8 c8}

) (

^{ a3+ +b3 c3}

)(

^{a5+ +b5 c5}

)

HD:

Ta có:

(

^{8 8}

) (

^{3 3}

)(

^{5 5}

)

2 a +b  a +b a +b

(

^{8 8}

) (

^{3 3}

)(

^{5 5}

)

2 b +c  b +c b +c

(

^{8 8}

) (

^{3 3}

)(

^{5 5}

)

2 c +a  a +c a +c Cộng theo vế ta được:

(

^{8 8 8}

) (

^{8 8 8}

) (

^{3 5 5 5}

) (

^{3 5 5 5}

) (

^{3 5 5 5}

)

4 a + +b c  a + +b c +a a + +b c +b a + +b c +c a + +b c

(

^{8 8 8}

) (

^{3 3 3}

)(

^{5 5 5}

)

3 a b c a b c a b c

= + +  + + + +

Bài 70: Cho a+b=2, CMR : a⁸+b⁸ a⁷+b⁷ HD:

Ta có: ²

(

^a8+b8

)

^

(

^{a b a+}

) (

^7+b7

)

^{=a8+ +b8 ab7+a b7}

= ^{a8+ −}