Sh. Ismailov, A. Qo’chqorov, B. Abdurahmoi

isbot. Teoremaning isboti quyidagi ma’lum tasdiqqa asoslangan: x> 1 da e^x-1>x , shu bilan birga e^x-1=x tenglik esa faqat x=1 da bajariladi. Bundan:

О a, A. a, ₁₄ ^A a, fG(a)^л"

1= e = exp \ ^-1| = Пexp^-:^ -1) >П

—

ⁱ
11^A(a) j i—1 ^A(^a) i—1^A(a) v^A(a) J

a

Demak, A(a) > G(a) va tenglik esa faqat —— — 1, i=1, 2,., n bo’lganda

A (a)

bajariladi. Bundan esa a₁=a₂ =... = a_n = A(a) ekanligi kelib chiqadi.

1. 
   misol. x, y > 0 bo’lsa, x² + y² +1 > xy + x + y tengsizlikni isbotlang.
   

2. 
   Yechilishi:2 2 2
   

2. 
   2 1 ^ x x y y
   

x + y +1 > xy + x + y ^ — + — + — + — +

2 2 2 2

1 1 2 2 i

+ +— — x + y + 1

>

xy,

>

+ <

2 22 2 x y —+— 2 2

y² 1


2 2

x² 1

+ — > x.

2 2


x² + y² +1 > xy + x + y.

2. 
   misol. x > 0 bo’lsa, 2^ + 2^ > 2* 2^ tengsizlikni isbotlang.
   

_L 1 1

Yechilishi. 2^Tx + 2^ > 2 * 2^^ * 2^fx — 2 * 2^x12+4 — 2 * 2^x6 — 2 * 2^Tx.

Misollar:

1. 
   Agar x, y > 0 bo’lsa, x¹ + y⁴ + 8 > 8xy ni isbotlang.
   
2. 
   x1, *^2 , x3, x4 , x5 0 bo’lsa quyidagini isbotlang:
   

( . . . \

x1 + x2 + x3 + x4 + x5 > x1 ( x2 + x3 + x4 + x5 ) .

3. 
   x, y, z > 0 bo’lsa, x² + y² + z² > xy + yz + xz ni isbotlang.
   

a b c

4. 
   a, b, c > 0 bo’lsa, —I 1— > 3 ni isbotlang.
   

b c a

a, b, c > 0 bo’lsa, (a +1) (b +1) (c + a) (b + c) > 16abc ni isbotlang

.(H(a)) ^_1= ^ai + ^a2 + ••• + ^an > ^ ¹a₂¹...a_n¹ = (G(a)) ¹ tenglikga ega

n

bo’lamiz. Jumladan, H(a) = G(a) tenglik faqat aj=a₂ = ... = a_nda bajariladi.

1. 
   r\ 3 a + b + c .......
   

1. 
   misol. Agar a, b, c > 0 bo lsa, —- - — < tengsizlikni
   

1/ a +1/ b +1/ c 3

isbotlang.

Yechilishi: 9 <(a + b + c)

a
> 9^abc = 9 ³ abc

'1 1 1^Л — + — + —

v a b c _J
+ b + c > 3-Vabc,

1111 ^(a + b + c)

• 
  + — + — > 3 • , . abc ³ abc
  

.23** 111

2. 
   misol. Agar a, b, c > 0, ab c — 1 bo’lsa, — + + — > 6 ni isbotlang.
   

abc

1 2 3 1 1 1 1 1 1 _r 1

Yechilishi: —\ \— — —\ \ \ \ \— > 6—. ~ — 6.

abc abbccc 6_ab²_c³

Misollar

1. 
   x, y, z > 0 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

1. 
   .1.1.1 > 28л/3
   

(x \ y \ z )^{2 x 2} y^{2 z 2} 9yj xyz (x \ y \ z )

2. 
   Agar x₁, x₂,...,x_n > 0 va x₁ \x₂ \...\ x_n — 1 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

^x1 , ^x2 , , ^xn > 1

V

^xn
^x1 ^{+ x}2 ⁺ ... ^{+ x}n yji¹ \ ^x1 )(^x2 \ .. \ ^xn ⁾ V(^{1 a x}1 ⁺ ... ⁺

3. 
   x, y, z > 0 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   
4. 
   
   

x² - xy y² - yz z² - xz _л

- + -—— + > 0.

x + y y + z x + z

5. 
   Agar a, b, c > 0 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

^a + Ll_+J^> 2.

b + c у a + c у a + b

6. 
   Agar a, b, c, d > 0 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

a^p+² + b^p+2 + c^p+² > a^p+²bc + b^p+2ac + c^p+2ab .

4. 
   O’rta arifmetik va o’rta kvadratik qiymatlar orasidagi tengsizlik.
   

Teorema. K (a) >A(a) tengsizlik o’rinli ekanligini, jumladan,

K(a) = A(a) tenglik faqat a₁=a₂ =... = a_n holdagina o’rinli bo’lishini isbotlang.

Isboti: Koshi tengsizligidan foydalanib (1-masalaga qarang) foydalanib

2a_ia_j < a² + a^, 1 < i < j < n tengsizlikni hosil qilamiz.

Demak,

(a₁ + a₂ +... + a) — a₁+ a₂ +... + a + 2 ^' aa.<