I-Way: abc is equivalent to 2.2.

Inequality problems

Ardak Mirzahmetov

Theorem 1. (

If then

Theorem 2.( and m k >0 then

Theorem 3.( Let be any real numbers, then

²

Theorem 4. ( then

Theorem 5. ( ,

Theorem 6.( If , ,

Theorem 7. ( If , then

Theorem 8. (Jensen For

CHAPTER - 1

1.1.(a,b ) a²+b²

1.2. (a,b>0) a+b

1.3. (a,b,c ) a²+b² +c² + bc + ca

1.4.(a,b,c ) (a+b+c)² 3( + bc + ca)

1.5. (a,b,c ) a²+b²+c²

1.6. (a,b,c 0) (a+b)(b+c)(c+a) 8abc

1.7. (a,b,c>0) (a+b+c)( ) 9

1.8. (a,b,c > 0)

1.9. (a,b,c 0) 6a+4b+5c 5

1.10. (a²+b²+c²=1) ab + bc + ca ≤ 1

1.11. (a,b ) a⁴ +b⁴ +2 4ab

1.12. (a,b,c 0) a³+b³+c³ + 6 3(a+b+c)

1.13. (n N and n>1)

1.14. (a,b,c ) a²+b²+c² +14

1.15. (a,b,c ) a⁴+b⁴+ ≥ ∙ abc

1.16. ≤ 4

1.17. (a,b,c ) 3a+4b+12c ≤ 13∙

1.18.

1.19. a²+b²+c²

1.20.

1.21. (a,b,c>0 , a+b+c=1)

1.22. a+b+c+abc

1.23. ad < bc

1.24. Compare:

1.25. Compare:

1.26.

1.27. (a,b,c are sides of a triangle) 2 (ab+bc+ca)

1.28. (x+y=2, x , y ≥ 0) 2

1.29. (x, y > 0)

1.30. (-1 < x , y, z < 1)

1.31. (a, b, c ≥ 0, a+b+c=1) 1+9abc ≥ 4

1.32. ( )

1.33. ( )

1.34. ( )

1.35.

1.36. Find the minimum value of

1.37. (a,b 0) a³+b³

CHAPTER - 2

2.1. ( )

2.2. (a,b,c 0) a³+b³+c³+3abc ab(a+b) +bc(b+c)+ ca(c+a)

2.3. (a,b,c>0) a² + b²+ c² + 2(ab+bc+ca)

2.4. (a,b>0)

2.5. (ab=1) a²+b² 2

2.6. (a,b,c 0) (a+b)(a+c) 2

2.7. (a,b,c,d>0) 4(a - d)

2.8. (a )

2.9. ( ,

(1+2 ) (1+2 ) … (1+2 )

2.10. (a,b,c>0 , a+b+c=1) (1+a)(1+b)(1+c)

2.11. (1+x ) (1+x)ⁿ

2.12. (a,b,c>0) ( )( )

2.13. (a,b,c>0 , abc=1 )

2.14. (0<a,b,c<1, a+b+c=2)

2.15. (a,b,c>0) abc

2.16. (a,b ) +

2.17. (a,b,c>0)

2.18. (a,b,c>0 , a+b+c=1) + +

2.19. (a,b,c>0 , abc=1 )

2.20. (a,b>0)

2.21. (a,b,c>0 , abc=1 ) + +

2.22. (x,y )

2.23. (x ) 3x³-6x²+4

2.24. (a,b>0, a+b=1)

2.25. (a,b,c ) a⁴+b⁴+c⁴ abc(a+b+c)

2.26. (n N and n>1, S_n=1+ ) n

2.27. ( a,b )

2.28. aⁿ-1 ≥ n∙

2.29. (a,b,c>0 , abc=1 ) + ≤

2.30. a+b >

2.31.

2.32.

2.33. abc

2.34. Find maximum and minimum values of z.

2.35.

2.36. (x+y+z >0) ≥

2.37. (a,b,c >0)

2.38. (a,b,c>0 , abc=1 )

2.39.

2.40.

2.41.

2.42.

2.43.

2.44.

2.45. (1-a)(1-b)(1-c)

2.46. (0<x,y,z<1) x(1-y)+y(1-z)+z(1-x)<1

2.47. (a, b, c > 0)

2.48. (x+y+z+t=0,

2.49.

2.50. (

2.51. x²+yz 2, y²+zx 2, z²+xy 2 ) Find maximum and minimum values of (x+y+z)

2.52.

2.53. (a,b,c are sides of a triangle)

2.54.

2.55.

SOLUTIONS

CHAPTER - 1

1.1. a²+b²

1.2. a+b

1.3. ⇔ a²+b² +c² + bc + ca

1.4. ⇔ (a+b+c)² 3( + bc + ca)

1.5. 3(a²+b²+c²) ≥ (a+b+c)² ⇔ (a- b)²+(b- c)²+(c- a)² ≥ 0

1.6.By 1.2.

1.7.ByA.M. G.M. : ,

(

1.8.A.M. G.M. :

1.9.A.M. G.M. : ⇔ ; ⇔ ; ⇔ ⇒

1.10. By 1.3. 1= a²+b² +c² + bc + ca

⇒

1.11. A.M. G.M. :

1.12. A.M. G.M. : ⇒

Similarly ⇒ Add them side by side.

1.13. A.M. G.M. :

1.14.(a-1)²+(b-2)²+(c-3)² ≥ 0 a²+b²+c²+14 ≥ 2(a+2b+3c)

1.15. By A. M. ≥ G.M.: a⁴+b⁴+ ≥ 4 ∙ ⁴ = ∙ abc

1.16. By Cauchy’s inequality: ² ≤ ≤(1²+1²+1²+1²)∙(a+b+c+d)=4∙8=32 ⇒ ≤ 4

1.17. By Cauchy’s inequality: (3a+4b+12c)²≤(3²+4²+12²)(a²+b²+c²)=

=169(a²+b²+c²) ⇒ 3a+4b+12c ≤ 13∙

1.18. By Cauchy’s inequality: (2x+3y)²=(2x+ ∙2y)² ≤ (2²+ )(x²+4y²)= ⇒

⇒ ≤ ⇒

1.19. By Cauchy’s inequality:

14²≤(a+2b+3c)²≤(a²+b²+c²)(1²+2²+3²)=(a²+b²+c²)14 ⇒ 14 ≤ a²+b²+c²

1.20. By Cauchy’s inequality:

( ∙ )² ≤ ( )( ) = ab

1.21.. By Cauchy’s inequality:( )² ≤ ≤(1²+1²+1²)( + ) =3∙(4(a+b+c)+3)=3∙7=21

1.22. By Cauchy’s inequality: (a+b+c)² ≤ (1²+1²+1²)(a²+b²+c²)=9 ⇒a+b+c ≤ 3;

By A. M. ≥ G.M.: 3 = a²+b²+c² ≥ 3∙³ =3∙ abc ≤ 1. Then a+b+c+abc ≤ 3+1=4

1.23. ad < bc ó ad < (a+d-c)c ó 0 < (c-d)(a-c)

1.24.For k ⇒ = <

<

1.25.1 , 2 ,

3 < …, 50

Multiply them side by side.

1.26. + +…+ > 2009∙ = .

1.27. By triangle inequality a+b > c, ⇒ ac+bc > c². Similarly ab+cb > b², ab+ac > a².

Add them side by side .

1.28. . Then given inequality is equivalent to

; Let’s prove it.

1.29.A.M-G.M.:

1.30.A.M-G.M.:

1.31.A.M-G.M.:

1.32. and ⇒ ⇒

1.33.

⇔

⇒

1.34. By A.M. G.M. , ⇒

1.35. By A.M. G.M.

⇒

1.36.A.M. G.M. :

1.37. a³+b³ ⇔ ⇔

⇔

CHAPTER - 2

2.1.By Cauchy’s inequality

2.2. By Schur’s inequality

⇔

a³+b³+c³+3abc ab(a+b) +bc(b+c)+ ca(c+a)

2.3.a²+b²+c²+ 2(ab+bc+ca) ⇔ Schur’s inequality

2.4. ⇔ ; . A.M. G.M. :

2.5.A.M. G.M. :

=

2.6.A.M. G.M : b

2.7.

⇒

⇔

2.8. ⇔

⇔ ⇔

⇔

2.9. , , …,

⇒

2.10. A.M. G.M. : =

⇒ , Similarly

; Multiply them side by side , then

2.11.If then

If , then by A.M. G.M. :

⇒ ⇒ (1+x)ⁿ

II-solution. By mathematical induction.

2.12. A.M. G.M. : ⇒

⇒ Similarly

; Multiply them side by side.

=1

2.13. abc=1 . ^: ⇒ , Similarly: ⇒

+ + 3a+3b+3c ⇒

2.14.1-a=x, 1-b=y, 1-c=z

. It is 1.6.

I-Way: abc is equivalent to 2.2.

II-way: Assume that a+b-c, b+c-a, c+a-b are positive and a+b-c=x , b+c-a=y, c+a-b=z It is 1.6.

2.16.By A.M. G.M. :

Similarly . +

2.17.A.M. G.M. :

Similarly

+ + + + = 2

Equality occurs when , But it is impossible .

2.18.. A.M . G.M.: Similarly

+

2.19. let a = x², b = y² , c = z², xyz = 1,

Let’s prove it. . A.M. G.M.: Similarly ⇒

( )+( )+( ) ⇒ (1)

AM GM: (2)

Add inequalities (1), (2) side by side.

2.20. + + >2 Is 2.17. . After substitutions a=x², b=y², c=1

it becomes

2.21. A.M. G.M. : ,

Similarly . Add them side by side,then . ⇒

Similarly .

2.22.

2.23. A. M. ≥ G.M. : x³+2x³+4 ≥ 3∙³ = 6x²

2.24. A. M. ≥ G.M.: = ⇒ ab ⇒ ≥ 4

By 2.1. + ≥ = ≥ =

2.25.Assume that a ≥ b ≥ c a³ ≥ b³ ≥ c³. By Chebyshev¢s inequality and

A.M. ∙ =

2.26. By A. M. ≥ G.M. : S_n+n=(1+1)+( +1)+…+( )=2+ + + … + ≥

n = n∙

2.27. By A. M. ≥ G.M. : a+ + ≥ 3∙ = 3∙ ,

b+ + ≥ 3∙ = 3∙ ⇒

ð a+b+4 ≥ 3( ) is equivalent to the given inequality

2.28. By A. M. ≥ G.M.: a^n-1+a^n-2+…+a+1 ≥ n∙ⁿ = =n∙ⁿ = n∙ⁿ = n∙

Multiply both sides by a-1, then aⁿ-1 ≥ n∙(a-1) = n∙

2.29. = = ≤ is true by A. M. ≥ H.M.

Similarly ≤ , ≤ . Then

+ ≤

2.30. a,b,k,n>0; ab>ka+nb ⇒ a- n > > 0; b-k > > 0 ⇒

a-n+b-k > + ≥ 2∙ = 2 ⇒ a-n+b- k >2

2.31. I-way. A. M. ≥ H.M.: [(a+b)+(b+c)+(c+a)]∙ ≥ 9 ó

≥ ó (

⇔ .

II- way: By 2.1.and 1.4.:

≥

2.32.By 1.37.

2.33. =x; =y; =z ⇒ x+y+z=1; a= = ;

Similarly: b= ; c= ; By 1.6. abc = ≥ 8

2.34. x+y=2-z; x²+y²=2-z²; by Cauchy’s inequality (x+y)² ≤ (x²+y²)(1²+1²) ⇒ (2-z)² ≤ 2(2-z²) ⇒ z(3z-4) ≤ 0 ⇒ 0 ≤ z ≤

2.35. By 2.1.: ≥ ≥ =1

2.36. By 2.1. and 1.3.:

≥ = ≥

2.37. The given inequality is equivalent to

≥

By Cauchy’s inequality and A. M. ≥ G.M.:

(1²+1²+1²) ≥ ²= ∙ ≥

∙3∙ ³ = 3 ≥ ;

A. M. ≥ G.M.: ≥ 3. Then ≥

2.38. By 2.1. and A. M. ≥ G.M.: =

= ≥ = ≥ =

2.39. ² ( )(1+2+…+n)=

=n(1+2+…+n) = . ⇒ .

By A.M H.M : ( + +…+ ) n² ⇒ ( + +…+ )

( + +…+ ) n²