1-misol
.
H
(
a
)
≥
min
{
a
1
, a
2
,…, a
n
}
va
max
{
a
1
, a
2
,…, a
n
}
≥
K
(
a
)
tengsizliklarni
isbotlang.
Yechimi:
Umumiylikni chegaralamagan holda
min{a
1
, a
2
,…, a
n
}= a
1
, max{a
1
, a
2
,…, a
n
}= a
n
deb hisoblash mumkin. U holda
10
H(a)=
1
1
1
2
...
n
n
a
a
a
−
−
−
+
+ +
1
≥
1
1
1
1
1
1
1
...
n
a
a
a
a
−
−
−
=
+
+ +
,
K
(
a
)
=
2
2
2
2
2
2
1
2
...
...
n
n
n
n
n
a
a
a
a
a
a
a
n
n
+
+ +
+
+ +
≤
=
bo’ladi.
Izoh 1
. yuqoridagi misollardan
max{a
1
, a
2
,…, a
n
}
≥
K(a)
≥
A(a)
≥
G(a)
≥
H (a)
≥
min{a
1
, a
2
,…, a
n
}
ekanligi kelib chiqadi
.
2-misol
.
(
)
(
)
2
2
2
2
3
a
b
c
a b c
+
+
≥
+ +
tengsizlikni isbotlang.
Yechilishi
:
(
) (
) (
)
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
2
2
2
0.
a
b
c
a
b
c ab
bc
ac
a
b
c
ab bc ac
a b
b c
c a
+
+
≥
+
+
+
+
⇒
⇒
+
+
≥
+
+
⇒ −
+ −
+ −
≥
3-misol
.
(
)(
)
(
) (
)
2
2
2
2
2
2
6
a
b
a
b
c
a b
a b c
+
+
+
≥
+
+ +
2
.
⇒
−
≥
tengsizlikni isbotlang.
Yechilishi:
(
)
(
)
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0
3
a
b
a b
a
b
a
b
ab
a b
a
b
c
a b c
⎧
+
≥
+
⎪
×
⇒
+
≥
+
+
⎨
+
+
≥
+ +
⎪⎩
(
)(
)
(
) (
)
2
2
2
2
2
2
2
6
a
b
a
b
c
a b
a b c
+
+
+
≥
+
+ +
.
Misollar
1. Agar
va
bo’lsa, u holda quyidagi tengsizlikni isbotlang:
, ,
0
a b c
>
1
a b c
+ + =
(
)
(
)
(
)
2
2
3
a
b c
b
a c
c
a b
+ −
+
+
−
+
+
−
≥
2
.
2. Agar
bo’lsa, u holda quyidagi tengsizlikni isbotlang:
, ,
0
a b c
>
(
)(
)(
)
3
3
3
2
2
2
8
4
a
b
c
abc
a b b c c a
a b b c a c
+
+
+
≥
+
+
+
+
+
.
11
3. Agar
bo’lsa, u holda quyidagi tengsizlikni isbotlang:
, ,
a b c R
∈
(
) (
) (
)
(
)
4
4
4
4
4
4
4
7
a b
b c
a c
a
b
c
+
+
+
+
+
≥
+
+
.
4. Agar
, ,
0
x y z
>
va
3
x y z
+ + =
bo’lsa, u holda quyidagi tengsizlikni isbotlang:
x
y
z
xy xz yz
+
+
≥
+
+
.
5. Agar
va
, ,
0
a b c
>
2
2
2
1
a
b
c
+
+
=
bo’lsa, u holda quyidagi tengsizlikni
isbotlang:
3
3
3
5
2
bc
ac
ab
a a
b b
c c
+
+
≥
−
−
−
.
§4. Umumlashgan Koshi tengsizligi.
Teorema. a
1
, a
2
,…, a
n
,
p
1
, p
2
,…, p
n
– musbat sonlar bo’lsin.
1
2
1
2
...
1 1
2
2
1
2
1
2
...
...
...
n
n
p p
p
p
p
p
n
n
n
n
a p
a p
a p
a a
a
p
p
p
+ + +
⎛
⎞
+
+ +
≤ ⎜
⎟
+
+ +
⎝
⎠
(1)
ekanligini isbotlang, tenglik esa faqat
a
1
= a
2
=…= a
n
da bajariladi.
Isboti:
s
=
1 1
2
2
1
2
...
...
n
n
n
a p
a p
a p
p
p
p
+
+ +
+
+ +
belgilash kiritamiz.
e
x -1
≥
x
(
x
≥
1)
tengsizlikka ko’ra
s
(
)
1
i
a
e
−
s
≥
a
i
, i=1, 2,…, n.
Bu tengsizliklarni barchasini ko’paytirib chiqamiz:
1
2
1
2
1
2
...
1 1
2
2
1
2
1
2
...
...
...
exp
...
.
n
n
n
p
p p
p
p
p
n
n
n
n
p p
p
a p
a p
a p
a a
a
s
p
p
p
s
s
+ + +
+ + +
+
+ +
⎛
⎞
≤
−
⎜
⎟
⎝
⎠
=
+
+ +
=
Tenglik faqat
s =a
1
=a
2
=…= a
n
da bajarilishi esa 1-masaladagidek
isbotlanadi.
12
1
2
1
2
...
1 1
2
2
1
2
1
2
...
...
...
n
n
p p
p
p
p
p
n
n
n
n
a p
a p
a p
a a
a
p
p
p
+ + +
⎛
⎞
+
+ +
≤ ⎜
⎟
+
+ +
⎝
⎠
Misol.
Quyidagi tengsizlikni isbotlang:
8
3
4
6
3 16 18
4
3
8
a
b
c
a b c
⎛
⎞
+
+
≥
⎜
⎟
⎝
⎠
.
Yechilishi:
Koshi tengsizligining umumiy holiga ko’ra
p
ning o’rnida 3
kelyapti.
1)
; ,
x R p q Q
∈
∈
bo’lsa,
(
)
2
2
2
sin
cos
p
q
p
q
p q
p q
x
x
p q
+
⋅
≤
+
ni isbotlang.
2)
12
2
3
4
6 12 20
3
4
5
12
a
b
c
a b c
⎛
⎞
+
+
≥
⎜
⎟
⎝
⎠
3)
bo’lsa,
, ,
0
a b c
>
2
2
2
2
2
2
2
2
2
2
4
7
a
b
c
ab bc ac
b
c
a
a
b
c
+
+
⎛
⎞
+
+
+
≥
⎜
+
+
⎝
⎠
⎟
isbotlang.
4)
bo’lsa,
, ,
0
a b c
>
2
2
2
3 1
a b
c
ab bc ac
b
c
a
a
b
c
+
+
+ + +
≥
+
+
+
isbotlang.
5)
bo’lsa,
, ,
0
a b c
>
2
2
2
2
6
a b b c a c
ab bc ac
c
a
b
a
b
c
+
+
+
+
+
2
+
+
+
≥
+
+
+
isbotlang.
§5. Umumlashgan Yung tengsizligi.
Teorema.
1
2
1
2
1 2
1
2
...
...
n
r
r
r
n
n
n
a
a
a
a a a
r
r
r
≤
+
+ +
(2)
13
tengsizlik urinli, bu yyerda
a
1
, a
2
,…, a
n
,
r
1
, r
2
, …, r
n
lar musbat sonlar, jumladan,
1
2
1
1
1
...
1
n
r
r
r
+ + +
=
.
Isboti:
5-masaladagi (1) tenglikda
a
i
ni
ga ,
r
i
r
i
a
i
ni esa
1
i
r
(
i=
1, 2, …,
n
) ga almashtirib
1
2
1
2
1 2
1
2
...
...
n
r
r
r
n
n
n
a
a
a
a a a
r
r
r
≤
+
+ +
ni olamiz.
Izoh
.
n=2
holida esa Yung klassik tengsizligiga ega bo’lamiz:
1
1
p
q
a
b
p
q
ab
+
≥
(
a
≥
0 , b
≥
0
)
,
(3)
bu yyerda
p, q
sonlar
1
1
1
p
q
+ =
tenglikni qanoatlantiruvchi musbat sonlar.
1-misol
. Agar
va
, ,
0
a b c
>
ab bc ac abc
+
+
=
bo’lsa,
b
c
a
b
c
abc
b
c
a
≤
+
+
a
Yechilishi:
Shartga ko’ra
1 1 1
1
ab bc ac abc
a b c
+
+
=
⇒ + + =
b
c
a
b
c
abc
b
c
a
≤
+
+
a
tengsizlik Yung tengsizligining xususiy holidan kelib
chiqadi.
2-misol.
Agar
bo’lsa,
, ,
0
a b c
>
2
3
6
18
12
6
36
a
b
c
abc
+
+
≥
ni isbotlang.
Yechilishi:
tengizlikni ikkala tomonini 36 ga
bo’lamiz
2
3
6
18
12
6
36
a
b
c
a
+
+
≥
bc
2
3
6
1
2
1
1
1
;
2
3
6
n
a
b
c
abc
r
r
r
+
+
≥
+
+ +
=
…
1
bo’lsa, Yung tengsizligi o’rinli.
2
3
6
1 1 1
1
2 3 6
2
3
6
a
b
c
abc
+ + = ⇒
+
+
≥
tengsizlik o’rinli.
14
Misollar
1)
Agar
, ,
0
x y z
≥
va
2
x y z
+ + =
bo’lsa,
3
3
3
3
3
3
2
x y y z z x
xy
yz
zx
+
+
+
+
+
≤
ni isbotlang.
2)
Agar
va
, ,
0
a b c
≥
2
2
2
3
a
b
c
+
+
=
bo’lsa,
(
)(
)(
)
2
2
2
ab
bc
ac
−
−
−
1
≥
ni isbotlang.
3)
Agar
va
, ,
0
a b c
≥
2
a b c
+ + =
bo’lsa,
2
2
2
2
2
2
3
a b b c c a
ab
bc
ca
+
+
+
+
+
≤
ni isbotlang.
4)
Agar
va
, ,
0
a b c
≥
1
a b c
+ + =
bo’lsa,
(
)
(
)
(
)
2
2
3
a
b c
b
c a
c
a b
+ −
+
+ −
+
+
−
≥
2
ni isbotlang.
5)
Agar
bo’lsa,
, ,
0
a b c
≥
4
a b b c a c
a
b
c
c
a
b
b c a c a b
+
+
+
⎛
⎞
+
+
≥
+
+
⎜
⎟
+
+
+
⎝
⎠
ni isbotlang.
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