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

V12 n J 12 n i j

1j

+ a_n + (a^ + a^^.) = n(a1 + a2 +... + a_n).

1j

Eslatib o’tamiz, K (a) = A(a) tenglik faqat a₁=a₂ =... = a_n o’rinli bo’ladi.

1-misol.

H (a) > min{a₁, a₂,..., a_n} va max{a₁, a₂,..., a_n} > K(a) tengsizliklarni

isbotlang.

Yechimi: Umumiylikni chegaralamagan holda

min{a₁, a₂,..., a_n}= a₁, max{a₁, a₂,..., a_n}= a_n deb hisoblash mumkin. U holdaI
_гг/ , " "

^H(a)= -1 . -1 .——1 > -1 . -1 .——г ^{= a}1


K(a)

a1 + a2 +.	.. + a n	> 1	a1 + a1 +.	.. + a1
1 2 2 j a1 + a2 +..	. + ^a2	• <	I^a2 + ^a" +..	. + ^a" —

n \ n

zoh 1. yuqoridagi misollardan

max{a₁, a₂,..., a_n} >K(a) >A(a) > G(a) >H (a) >min{a₁, a₂,..., a_n} ekanligi kelib chiqadi.

2. 
   misol. 3 (a² + b² + c²) > (a + b + c)² tengsizlikni isbotlang.
   

Yechilishi:

3a + 3b + 3c > a + b + c 2ab + 2bc + 2ac ^ a² + b² + c² > ab + bc + ac ^(a - b)² +(b - c)² +(c - a)² > 0.

3. 
   misol. 6 (a² + b ² )a² + b² + c²) >( a + b )² (a + b + c )² tengsizlikni isbotlang.
   

Yechilishi:

2. 
   (a² + b² )(a + b )² ₂
   

x< ^ 2a² + 2b² > a² + b² + 2ab ^ (a -b) >0.

3. 
   (a² + b² + c¹) > (a + b + c )²
   

6. 
   (² + b²) ² + b² + c²) > (a + b)² (a + b + c)².
   

Misollar

1. 
   Agar a, b, c > 0 va a + b + c — 1 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

■\Ja + (b-c)² +yjb + (a-c)² +^c + (a-b)² ^V3.

2. 
   Agar a, b, c > 0 bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

a³ + b³ + c³ 8abc

3. 
   a ²b + b² c + c² a (a + b )b + c )(a + c)Agar a, b, c e R bo’lsa, u holda quyidagi tengsizlikni isbotlang:
   

(a \ b)⁴ \ (b \ c)⁴ \ (a \ c)⁴ > -4- (a⁴ \ b⁴ \ c⁴).

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

л/x \ -yjy \ yfz > xy \ xz \ yz.

5. 
   Agar a, b, c > 0 va a² \ b² \ c² — 1 bo’lsa, u holda quyidagi tengsizlikni
   

isbotlang:

bc ac ab 5

\ r\ > .

a - a b - b c - c 2

§4. Umumlashgan Koshi tengsizligi.

T
(1)
eorema. a₁, a₂a_n,p₁, p₂p_n - musbat sonlar bo’lsin. ap¹ ap²...a^Pn <

12 n

v P1 \ P2 \ ... \ Pn J

ekanligini isbotlang, tenglik esa faqat a₁= a₂ =... = a_n da bajariladi.

a,p, \a₂p₂ \...\a p

Isboti: s =—^ belgilash kiritamiz.

p_x \ p₂ \... \ p_n

e^x-1 > x ( x>1) tengsizlikka ko’ra s e^(a‘ ¹^^s> a_t, i=1, 2,..., n.

Bu tengsizliklarni barchasini ko’paytirib chiqamiz:
a*a^p2 ...a^pn < s^p1 -^p2^^pn

12 n

p1 \ p₂\...\pn

■p1+ p2⁺- + \n

exp

s

T
an da bajarilishi esa 1-masaladagidek
englik faqat s =a₁=a₂ isbotlanadi

./ \p1+Pi+...+Pn

^a1A + ^a2 p2 + ... + ^a,pn

af¹ a2p²...a^Pn <

12 n

p_x + p 2 + ... + p, J

Misol. Quyidagi tengsizlikni isbotlang:

v

л⁸

3

> a³b¹⁶ c¹⁸.

a² + 4b⁴ + 3c

8

Yechilishi: Koshi tengsizligining umumiy holiga ko’ra p ning o’rnida 3 kelyapti.

p q

p² q²

1. 
   x
   p+q
   
   2
   e R; p,q eQbo’lsa, sin^p x*cos^q x < ni isbotlang.
   

(p+q

12

2)

²

2 i 2 2 a b c

~r + T + —; b c a

ab + bc + ac

3) a, b, c > 0 bo’lsa,

> 7 isbotlang.

)f 3a² + 4b³ + 5c 12


> a ⁶b¹² c²⁰


_v a² + b² + c² у

>л/б+ 2

5) a, b, c > 0bo’lsa, isbotlang.

b

a

a + b b + c a + c _л lab + bc + ac + + + 2,

a² + b² + c²

§5. Umumlashgan Yung tengsizligi.

Teorema.

a¹} a2 a^r" a₁ a₂...a < — + — +... + —— (2)

^r1 ^r2 ^r

ntengsizlik urinli, bu yyerda a₁, a₂,..., a_n, r₁, r₂, ..., r_n lar musbat sonlar, jumladan,

11 1 _—!

—\ \... \— — 1.

^Г1 ^Г2 ^rn

Isboti: 5-masaladagi (1) tenglikda a_t ni a^r' ga , r ni esa 1/r (i=1, 2, ..., n) ga almashtirib

a^r1 a^r2 a^rn a₁ a₂...a_n <\...ni olamiz.

^r1 ^r2 ^rn

Izoh. n=2 holida esa Yung klassik tengsizligiga ega bo’lamiz:

—a^p \¹ b^q > ab ( a> 0 , b> 0), (3)

p q

bu yyerda p, q sonlar — \¹ — 1 tenglikni qanoatlantiruvchi musbat sonlar.

p q

_a^b b^c _c^a

1. 
   misol. Agar a, b, c > 0 va ab \ bc \ ac — abc bo’lsa, abc < \ \
   

b c a

Yechilishi: Shartga ko’ra ab \ bc \ ac — abc ^\\¹ — 1

abc

_a^b b^c _c^a

abc < \ \ tengsizlik Yung tengsizligining xususiy holidan kelib

b c a

chiqadi.

2. 
   misol. Agar a, b, c > 0 bo’lsa, 18a³ \ 12b⁴ \ 6c⁶ > 36abc ni isbotlang.
   

Yechilishi: 18a² \ 12b³ \ 6c⁶ > 36abc tengizlikni ikkala tom on ini 36 ga

a² b³ _c⁶ 11 1

bo’lamiz \ \ > abc;—\ \ ... \— — 1 bo’lsa, Yung tengsizligi o’rinli.

2 3 6 r r₂ r