§6. Gel’der tengsizligi.
Teorema.
1
1
1
p q
+ =
shartni qanoatlantiruvchi barcha musbat
p
,
q
sonlar va
a
j
,
b
j
,
j
= 1, ...,
n
sonlar uchun
1
1
1
1
1
|
|
| |
n
n
n
p
q
p
i i
i
i
i
i
i
a b
a
b
=
=
=
⎛
⎞ ⎛
≤ ⎜
⎟ ⎜
⎝
⎠ ⎝
∑
∑
∑
q
⎞
⎟
⎠
q
≠
(4)
tengsizlik har doim to’g’ri.
Isboti.
deb faraz qilamiz (aks holda (4) tengsizlik
bajarilishi ravshan). Yung tengsizligini qo’llab
1
1
|
|
0,
| |
0
n
n
p
i
i
i
i
a
b
=
=
≠
∑
∑
15
1
1
1
1
1
|
|
|
|
n
i
i
i
n
n
p
q
p
q
k
k
k
k
a
b
a
b
=
=
=
⎛
⎞ ⎛
⎜
⎟ ⎜
⎝
⎠ ⎝
∑
∑
∑
⎞
⎟
⎠
≤
1
1
1
1
1
|
|
| |
|
|
|
|
n
i
i
i
n
n
p
q
p
q
k
k
k
k
a
b
a
b
=
=
=
⎛
⎞ ⎛
⎜
⎟ ⎜
⎝
⎠ ⎝
∑
∑
∑
⎞
⎟
⎠
≤
≤
1
1
1
|
|
| |
|
|
|
|
p
q
n
i
i
n
n
p
q
i
k
k
k
k
a
b
p
a
q
b
=
=
=
∑
∑
∑
=
1
1
1
p
q
+ =
ga ega bo’lamiz. Bu yyerdan (4) tengsizlik kelib chiqadi.
Izoh
.
Gyol’der tengsizligining
p
=
q
= 2 dagi
2
2
1
1
n
n
n
i i
i
i
i
i
i
a b
a
b
=
=
=
≤
1
∑
∑
∑
Koshi-Bunyakovskiy-Shvarts
tengsizligi deb ataluvchi bir muhim hususiy holini
aytib o’tamiz.
1-misol
(Minkovskiy tengsizligi). Ixtiyoriy musbat
a
j
,
b
j
(
j
= 1,...,
n
) sonlar va
natural
p
son uchun
(
)
≤
1/
(
)
p
p
k
k
a
b
+
∑
(
)
1/
p
p
k
a
∑
(
)
1/
p
k
b
p
∑
(5)
tengsizlikni isbotlang
Yechilishi.
(a
k
+b
k
)
p
= a
k
(a
k
+b
k
)
p-1
+ b
k
(a
k
+b
k
)
p-1
(
k=1, 2, …, n
) tengsizlikni
qo’shib,
∑
(a
k
+b
k
)
p
=
∑
a
k
(a
k
+b
k
)
p-1
+
∑
b
k
(a
k
+b
k
)
p -1
ni olamiz.
(4) tengsizlikka ko’ra
∑
a
k
(a
k
+b
k
)
p-1
≤
(
)
1/
p
p
k
a
∑
(
)
(
1)
1/
(
)
q p
q
k
k
a
b
−
+
∑
,
∑
b
k
(a
k
+b
k
)
p-1
≤
(
)
1/
p
p
k
b
∑
(
)
(
1)
1/
(
)
q p
q
k
k
a
b
−
+
∑
larga ega bo’lamiz, bu yyerdan
q
(
p-
1)
= p
tenglik yordamida (6) tengsizlik kelib
chiqadi.
16
Misollar
1)
(
)
(
) (
)
1
1
2
2
2
2
2
1 1
2 2
1
2
1
2
a b
a b
a
a
b
b
+
≤
+
⋅
+
2
Gyol’der tengsizligining
2
p q
= =
holiga ko’ra o’rinli.
2)
(
)
5
5
5
5
5
5
5
5
2
2
3
3
3
3
3
3
1 1
2 2
3 3
1
2
3
1
2
3
a b
a b
a b
a
a
a
b
b
b
⎛
⎞ ⎛
+
+
≤
+
+
⋅
+
+
⎜
⎟ ⎜
⎝
⎠ ⎝
⎞
⎟
⎠
Gyol’der
tengsizligining
5
3,
,
3
2
n
p
q
5
=
=
=
holiga ko’ra o’rinli.
3)
Agar
va
, ,
0
a b c
≥
3
3
3
3
a
b
c
+
+
=
bo’lsa,
ni
isbotlang.
4 4
4 4
4 4
3
a b
b c
c a
+
+
≤
4)
Agar
bo’lsa,
isbotlang.
, ,
0
a b c
≥
(
)
2
2
2
2
1 2
a
b
c
abc
ab bc ac
+
+
+
+ ≥
+
+
5)
Agar
va
, ,
0
a b c
≥
1
abc
=
bo’lsa,
2
2
5
3
3
a b c
a
b
c
2
+ +
+
+
≥
ni
isbotlang.
Amaliyot uchun masalalar.
1-masala.
Tengsizliklarni isbotlang:
!
,
3
n
n
n
n
⎛ ⎞
> ⎜ ⎟
⎝ ⎠
N
∈
; (1)
1
!
,
:
2
n
n
n
n N
+
⎛
⎞
<
∀ ∈
⎜
⎟
⎝
⎠
2
n
≥
; (2)
2
!
,
:
n
n
n
n N n
>
∀ ∈
≥
3
n
≥
; (3)
; (4)
1
! 2 ,
:
3
n
n
n N
−
>
∀ ∈
17
,
2
n
n
n
n
n e
n N
e
⎛ ⎞
⎛ ⎞
< <
∀ ∈
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
. (5)
(1) tengsizlikni isbotlaymiz.
Induktsiya bazasi
.
1
n
=
da:
1
1
1!
3
⎛ ⎞
> ⎜ ⎟
⎝ ⎠
ga egamiz. Induktsiya bazasi
isbotlandi.
Induktiv o’tish.
da
n k
=
!
3
k
k
k
⎛ ⎞
> ⎜ ⎟
⎝ ⎠
tengsizlik to’g’ri deb faraz qilamiz.
da tengsizlik bajarilishini isbotlaymiz:
1
n k
= +
(
1)
(
1)
(
1)!
3
k
k
k
+
+
⎛
⎞
+
> ⎜
⎟
⎝
⎠
.
(
1)!
!(
1)
(
1
3
k
k
k
k k
k
⎛ ⎞
+
=
+ >
+
⎜ ⎟
⎝ ⎠
)
ga egamiz.
(
1)
1
3
k
k
+
+
⎛
⎞
⎜
⎟
⎝
⎠
songa ko’paytiramiz va bo’lamiz:
1
(
1)
(
1)
1
3
(
1)
3
(
1)
3
k
k
k
k
k
k
k
k
k
+
+
+
+
⋅
⋅ +
⎛
⎞
=
⎜
⎟
+
⋅
⎝
⎠
1
1
1
3
1
1
3
3
(1
)
k
k
k
k
k
k
+
+
+
+
⎛
⎞
⎛
⎞
=
>
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
+
.
Bu yyerda quyidagi joriy hisoblashlarni bajaramiz:
2
1
1
(
1) 1
(
1) ... (
1) 1
(1
)
1
...
2!
!
k
k
k k
k k
k k
k
k
k
k
k
−
− ⋅ ⋅ − +
+
= + +
⋅
+ +
⋅
=
1
1
1
1
1
1
2
1
1 1
1
...
1
1
... 1
2!
!
k
k
k
k
k
k
<
<
−
⎛
⎞
⎛
⎞⎛
⎞
⎛
⎞
= + +
−
+ +
−
−
⋅ ⋅ −
<
⎜
⎟
⎜
⎟⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠⎝
⎠
⎝
⎠
18
2
1
2
1
1
1
1
1
2 3 2
1 2 3 ...
2
1
1
1
1
1
1
1 1
...
1 1
...
2!
3!
!
2 2
2
k
k
k
k
−
−
=
<
=
<
⋅
⋅ ⋅ ⋅ ⋅
= + +
+
+ +
< + + +
+ +
<
2
1
1
1
1
1
1
1
3
1 1
...
... 1
3
(1
)
3
1.
1
1
2 2
2
2
1
(
2
k
k
k
k
k
k
−
< + + +
+ +
+
+ < +
= ⇒ +
< ⇒
>
−
+
1
)
Matematik induktsiya printsipiga asoslanib, ixtiyoriy natural son uchun (1)
tengsizlik bajariladi deb xulosa qilamiz.
n
(2) tengsizlikni isbotlaymiz.
Induktsiya bazasi
.
2
n
=
da:
(2) tengsizlikning chap tomoni:
2! 2
=
;
(2) tengsizlikning o’ng tomoni:
2
2
2 1
3
9
2, 25
2
2
4
+
⎛
⎞
⎛ ⎞
=
= =
⎜
⎟
⎜ ⎟
⎝
⎠
⎝ ⎠
.
2 2
demak,
, 25
<
induktsiya bazasi isbot bo’ldi.
Induktiv o’tish.
da
n k
=
1
!
,
2
k
k
k
+
⎛
⎞
2
k
<
≥
⎜
⎟
⎝
⎠
tengsizlik to’g’ri deb faraz
qilamiz.
da
1
n k
= +
1
2)
(
1)!
,
3
k
k
k
+
+
⎛
⎞
+
<
≥
⎜
⎟
⎝
⎠
2
k
tengsizlik bajarilishini isbotlash
kerak.
1
(
1)!
! (
1)
(
1)
2
k
k
k
k k
k
+
⎛
⎞
+
= ⋅ + <
⋅ +
⎜
⎟
⎝
⎠
=
ga egamiz.
1
2
2
k
k
+
+
⎛
⎞
⎜
⎟
⎝
⎠
songa
ko’paytiramiz va bo’lamiz:
1
1
1
1
1
1
2
(
1) (
1) 2
2
2 (
1)
2
2
2 (
2)
(
2)
k
k
k
k
k
<
k
k
k
k
k
k
k
k
k
k
+
+
+
+
+
+
+
+
⋅ + ⋅
+
⋅ +
⎛
⎞
⎛
⎞
=
=
⎜
⎟
⎜
⎟
⋅ +
+
⎝
⎠
⎝
⎠
1
1
2 (
1)
1
(
2)
k
k
k
k
+
+
⋅ +
<
+
tengsizlik bajarilishini isbotlaymiz.
19
1
1
1
2 (
1)
2
1
2
(
2)
2
1
1
1
1
k
k
k
k
k
k
k
k
1
k
+
+
+
+
⋅ +
=
= ⋅
+
+
⎛
⎞
⎛
⎞
+
⎜
⎟
⎜
⎟
+
+
⎝
⎠
⎝
⎠
1
2
0
1
1 (
1)
1
1
(1
)
1
...
2
1
1
2!
1
(
1)
k
k
k
k
k
k
k
k
k
+
>
+
+ ⋅
⎛
⎞
+
= +
+
⋅
+ + ⎜
⎟
+
+
+
+
⎝
⎠
>
.
1
1
1
1
2
1
(1
)
2
2
1
1
2
1
2
k
k
k
k
k
+
+
+
⎛
⎞
⇒ +
<
⇒ ⋅
< ⋅ =
⎜
⎟
+
+
⎝
⎠
.
1
1
2
2
1
2
2
k
k
k
k
+
+
+
+
⎛
⎞
⎛
⎞
<
⋅ =
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
.
Matematik induktsiya printsipiga asoslanib, ixtiyoriy
natural son uchun
(2) tengsizlik bajariladi deb xulosa qilamiz.
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