Sonli tengsizliklar haqida. Toshkent 2008 - bet 21

6.
Berilgan tengsizlikni chap tomonini
T
bilan belgilab, o’rta arifmetik va o’rta
geometrik miqdorlar haqidagi Koshi tengsizligidan foydalanamiz, ya’ni:
1
1
1
(1
)(1
)
(1
)(1
)
(1
)(1
)
1
1
1
2 1
1
2 1
1
2 1
1
1
3
2
2
ab
bc
ac
T
ab
a b
bc
b c
ac
a c
ab
bc
ac
a
b
b
c
a
c
a
b
b
c
a
c
b
a
c
b
c
a
a
b
b
c
a
c
a c b c b a
c a b a b c
=
+
+
=
+ − −
+ − −
+ − −
=
+
+
≤
−
−
−
−
−
−
⎛
⎞
⎛
⎞
⎛
⎞
≤
+
+
+
+
+
≤
⎜
⎟
⎜
⎟
⎜
⎟
−
−
−
−
−
−
⎝
⎠
⎝
⎠
⎝
⎠
⎛
⎞
≤
+
+
+
+
+
=
⎜
⎟
+
+
+
+
+
+
⎝
⎠
7.
Tengsizlikni ikkala qismidagi qavslarni ochib ixchamlasak
2
2
2
2
2
2
3
a
b
c
a
c
b
a b
c
b
c
a
c
b
a
b
c
a
+
+
+ + + ≥ + + +
(*)
ifoda hosil bo’ladi.Endi
,
,
a
b
c
x
y
z
b
c
a
=
=
=
deb belgilash kiritsak, u holda (*)
tengsizlik
2
2
2
1
1
1
3
x
y
z
x y z
x
y
z
+
+
+ + + ≥ + + +
(**)
ko’rinishga keladi.
1
xyz
=
ekanligidan quyidagi
(
)
2
2
2
2
3
x y z
x
y
z
x y z
+ +
+
+
≥
≥ + +
(1)
va
1
1
1
3
x
y
z
+ + ≥
(2) tengsizliklar o’rinli. (1) va (2) larni hadma-had qo’shib (**) ni
hosil qilamiz. Bundan (*) isbotlandi.
8.
1
a b
+ =
ekanligidan foydalanib yuqoridagi tengsizlikni quyidagi
23
shaklda yozamiz:
(
)
(
)
(
)
(
)
(
)
(
)
2
2
1
3
a
b
a b a
a b
a b b
a b
≤
+
+
+
+
+
+
+
Bundan
2
2
3
3
a b ab
a
b
+
≤
+
yoki
(
) (
)
(
) (
)
2
3
3
2
2
0
a
b
a b ab
a b
a b
+
−
+
=
−
+
≥
9.
2
2
3
3
4
4
(
)
yz y
z
y z yz
y
z
+
=
+
≤
+
tengsizlik o’rinli, chunki
(
)
(
)
4
3
3
4
3
3
4
4
2
2
2
2
(
) 0
(
)
y
y z yz
z
y
z
y z
x y
z
xyz y
z
y
z
−
−
+
=
−
− ≥ ⇒
+
≥
+
≥
+
yoki
5
5
4
5
2
2
5
4
4
4
4
4
(
)
x
x
x
x
y
z
x
x y
z
x
y
z
≥
=
+
+
+
+
+
+
.
Xuddi shunday,
5
4
5
4
5
2
2
4
4
4
5
2
2
4
4
4
,
y
y
z
z
y
x
z
x
y
z
z
x
y
x
y
z
≥
≥
+
+
+
+
+
+
+
+
.
Bu tengsizliklarni hadma-had ko’shib, isboti talab qilingan tengsizlikni hosil
qilamiz.
10.
Yuqoridagi tengsizlikni quyidagicha yozib olamiz:
2
2
2
2
2
2
2
2
2
5
2
2
5
2
2
5
2
2
3
x
y
z
y
x
z
x
y
z
x
y
z
x
z
x
z
x
y
+
+
+
+
+
+
+
+
≤
+
+
+
+
+
+
va Koshi-Bunyakovskiy-Shvarts tengsizligini qo’llab,
5
1
5
2
2
2
2
2
2
2
2
2
2 2
2
2
(
)(
) ( ( )
)
(
)
x
y
z
yz y
z
x yz
y
z
x
y
z
+
+
+
+
≥
+
+
≥
+
+
yoki
2
2
2
2
2
5
2
2
2
2
2
.
x
y
z
yz y
z
x
y
z
x
y
z
+
+
+
+
≤
+
+
+
+
Xuddi shunday,
2
2
2
2
2
2
2
2
2
2
5
2
2
2
2
2
5
2
2
2
2
2
,
x
y
z
xz x
z
x
y
z
xy x
y
x
y
z
x
y
z
z
x
y
x
y
z
+
+
+
+
+
+
+
+
≤
≤
+
+
+
+
+
+
+
+
munosabatlarni hosil qilamiz. Bu tengsizliklarni hadma-had qo’shsak,
2
2
2
2
2
2
2
2
2
5
2
2
5
2
2
5
2
2
2
2
2
2
3
x
y
z
x
y
z
x
y
z
xy yz zx
x
y
z
y
x
z
z
x
y
x
y
z
+
+
+
+
+
+
+
+
+
+
≤ +
≤
+
+
+
+
+
+
+
+
24
11
. , ,
α β γ
uchburchak burchaklari uchun
2
2
2
1 cos
cos
cos
2cos cos cos
α
β
γ
α
β
γ
=
+
+
+
tenglikdan foydalanib,
cos
cos
cos
cos
cos
cos
2
cos cos
cos cos
cos cos
cos cos
cos cos
cos cos
α
β
γ
α
β
γ
β
γ
γ
α
α
β
β
γ
γ
α
α
β
⋅
⋅
=
+
+
+
ifodani hosil qilamiz .
cos
cos
cos
,
,
cos cos
cos cos
cos cos
x
y
z
α
β
γ
β
γ
γ
α
α
β
=
=
=
deb belgilash kiritib,
3
cos
cos
cos
2
α
β
γ
+
+
≤
tengsizlikdan foydalansak,
1
1
1
3
2(
) 3
2
4(
2(
)) 9
8(
) 9(
2) 4(
) 5(
) 18
x
y
z
xyz
xy
yz
zx
x y z
xy
yz
zx
xyz
xy
yz
zx
x y z
x y z
x y z
+
≤ ⇔
+
+
≤
⇔
+
+ + +
+
+
≤
⇔
+
+
≤
+ + +
−
+ +
=
+ + +
12.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan
foydalanib,
(
)
(
)
(
)
(
)
(
)
(
)
2
2
3
2
3
2
2
3
3
2
2
1
1
1
, 1
1
1
,
2
2
2
1
1
1
2
a
b
a
a
a
a
b
b
b b
c
c
c
c c
+
+
+
=
+
+
−
≤
+
=
+
− +
≤
+
+
=
+
− +
≤
munosabatlarni topamiz. Endi quyidagi tengsizlikni isbotlasak yetarli:
(
)(
) (
)(
) (
)(
)
(
) (
) (
)
(
)
(
)(
)(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2 2
2 2
2 2
2
2
2
2 2 2
4
4
4
4
3
2
2
2
2
2
2
3
2
2
2
2
2
2
2
8 72
a
b
c
a
b
b
c
c
a
a
c
b
a
c
b
a
b
c
a b
b c
c a
a
b
c
a b c
+
+
≥ ⇔
+
+
+
+
+
+
+
+
+
+
+
≥
+
+
+
⇔
+
+
+
+
+
≥
+ =
Bu tengsizlik quyidagi tengsizliklarni hadma-had qo’shishdan hosil qilinadi:
(
)
(
)
4
2 2
2 2
2 2
2
2
2
2 2 2
3
3
3
48, 2
6
24
a b
b c
c a
abc
a
b
c
a b c
+
+
≥
=
+
+
≥
=
Bulardan isboti talab qilingan tengsizlikni hosil qilamiz.
25
13.
Birinchi navbatda
2
2
3
x y
x
y
+
≥
+
tengsizlikni isbotlaymiz. Faraz qilaylik,
2
2
3
x y
x
y
+
<
+
bo’lsin.
3
4
2
3
x
y
x
y
+
≤
+
tengsizlikdan foydalansak, farazimizga
zid bo’lgan
(
) (
) (
)
2
3
3
2
4
2
3
2
2
2
x
y
x x
y
y
x
y
+
≥
+
+
+
≥
+
tengsizlik hosil bo’ladi.
Shuning uchun
(
)
2
2
3
3
4
2
2
3
3
4
3
3
2
3
3
2
2
1 2
1
2.
x y
x
y
x
y
x y
x
y
x
y
x
y
x
y
x
y
+
≥
+
≥
+
⇒
+
≥
+
+
+
≥
≥
+
+
− +
− ⇒
+
≤
14.
Umumiylikni chegaralamasdan
a b c
≥ ≥
deb olib, uchburchak tengsiz-ligini
qo’llasak,
1
1
2
2
a b c
a
b a
= + + >
⇒ ≤ <
va bundan
(
)
(*)
2
2
2
2
2
1
2
1
1
1
n
n
n
n
n
n
n
n
n
b
a
b
a
<
+
⇒
=
+
<
+
.
Endi qo’yidagini qaraymiz:
1
....
2
2
2
n
n
n
n
n
n
n
c
n
c
b
b
c b
b
c
−
⎛
⎞
+
=
+
+
+
>
+
⎜
⎟
⎝
⎠
(chunki
1
2
n
n
n
cb
c
−
>
).
Xuddi shunday,
2
n
n
n
c
a
a
c
⎛
⎞
+
>
+
⎜
⎟
⎝
⎠
.
Demak,
(
) (
)
1
1
1
2
2
n
n
n
n
n
n
c
c
b
c
a
b
b
a
+
+
+
< + + + =
. (**)
(*) va (**) larni hadma-had qo’shib, isboti talab etilgan tengsizlikni hosil qilamiz.
15.
1
k
k
k
A
x
A
−
=
va
1
1
x
=
deb belgilash kiritsak,
2
2
2
1
3
1 2
1
2
3
1
2
3
4
...
...
n
n
n
n
n
n
n
n
n
n
n
G
A A A
A A
A
A
n
n
n
A
A
A
A
A
A
−
−
⎛
⎞
⎛
⎞
⎛
⎞
= ⋅
= ⋅
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
26
(
)
2
2
(
1)
2 3
1
2
3
1
2
2 3 4
1
2
3
4
1
1
...
....
;
1
(
1)
(
1) ;
n n
n
n
n
n
n
n
k
k
k
k
k
k
k
k
n x x x
x
n x
x x x
x
kA
k
A
a
A
k
k
k
k
x
A
A
A
+
−
−
−
−
=
=
⋅
−
−
=
= −
−
= −
−
1
2
2
3
1
2
...
1.(2
)(3 2 )...(
(
1) )
n
n
n
n
k
n
n
g
a a
a
x
x
n
n
x
G
A A
A
=
⋅
=
−
−
−
−
.
Umumlashgan Koshi tengsizligidan foydalansak
(
0,
0,
1,2,...,
i
i
a
i
n
α
>
>
=
1
2
1
2
1 1
2
2
1
2
1
2
...
...
...
n
n
n
n
n
n
a
a
a
a a
a
α
α
α
α α α
α
α
α
α α
α
+ +
+
+ +
≤
+
+ +
),
2
(
1)
2
1
2
1
2 3
2
3
2
3
2
3
...
1 (2
)(3 2 )...(
(
1) )
1
(
1)
1
2
... (
1)
(1 (2
) (3 2 ) ... (
1) )
2
n n
n
n
n
n
n
n
n
n
n
n
n
n
G
g
n
n x
x x
x
x
x
n
n
x
A
G
n n
x
x
n
x
x
x
n
x
n
n
+
−
+
=
+
⋅ −
−
−
−
≤
+
⎛
⎞
≤
+
+
+ +
−
+
+
−
+ −
+ +
−
⎜
⎟
⎝
⎠
2
3
2
3
1 1
(
1) 1
(
2
... (
1) )
(
2
... (
1) )
1
2
2
n
n
n
n
x
x
n
x
x
x
n
x
n
n
n
+
+
+
+
+ +
−
+
−
+
+ +
−
= +
.
16.
Bu tengsizlikning chap tomonini
S
deb belgilab, quyidagi usulda o’rta
arifmetik va o’rta geometrik miqdorlar o’rtasidagi munosabatni qo’llaymiz:
1 1 (1
)
1 1 (1
)
1 1 (1
)
3
3
3
3
3
3
1
3
b c
c a
a b
S a
b
c
a
b
c ab ac bc ba ca cb
+ + + −
+ + + −
+ + + −
⎛
⎞
⎛
⎞
⎛
⎞
≤
+
+
=
⎜
⎟
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎝
⎠
+
+
+
+
+
−
−
−
=
17.
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
(1
)
1
(1
)
ab
bc
cd
da
ab
cd
bc
da
a
b
c
d
a
c
b
d
ab
ab
bc
bc
a
ab
c
b
bc
d
+
+
+
+
+
+
+
+
⎛
⎞ ⎛
⎞
+
+
+
=
+
+
+
=
⎜
⎟ ⎜
⎟
+
+
+
+
+
+
+
+
⎝
⎠ ⎝
⎠
⎛
⎞ ⎛
⎞
+
+
+
+
=
+
+
+
=
⎜
⎟ ⎜
⎟
+
+
+
+
⎝
⎠ ⎝
⎠
(
)
1
1
1
1
(1
)
(1
)
1
(1
)
1
(1
)
4 1
4(1
)
4
1
(1
) 1
(1
)
ab
bc
a
ab
c
b bc
d
ab
bc
a ab
c
b bc
d
⎛
⎞
⎛
⎞
= +
+
+ +
+
≥
⎜
⎟
⎜
⎟
+
+
+
+
⎝
⎠
⎝
⎠
+
+
≥
+
=
+ +
+
+ +
+
27
18.
Bu tengsizlikni chap tomonini
T
bilan belgilab, umumlashgan Koshi-
Bunyakovskiy-Shvarts tengsizligini quyidagicha qo’llaymiz:
(
)
(
)
3
1
(
1)
(
1)
(
)
1
T a b
b c
c a
a b c
⋅
+ +
+ +
+
≥
+ +
=
yoki
1
1
T
ab bc ca
≥
+
+
+
tengsizlikni va undan
2
1
1
3
(
)
1
4
1
3
T
a b
с
ab bc ca
≥
≥
=
+ +
+
+
+
+
munosabatni hosil qilamiz.
-19.
2
2
2
3
(1
)
(1
)
(1
)
1
3
1 2
a
b b
c c
a
a
b c
b b
a c
c c
a b
a
b c
c a
a b
b c
c a
a b
a b
b c
c a
a b b c
c a
a
b
c
b c
c a
c b
b c
c a
a b
a b b c c a
b c c a a b
+
+
+
− − +
− − +
− − +
+
+
=
+
+
=
+
+
+
+
+
+
+
+
+
+
+
+
=
− +
− +
− =
+
+
− ≥
+
+
+
+
+
+
+
+
+
≥
⋅
⋅
− =
+
+
+
20.
Ixtiyoriy , ,
x y z
>0 uchun
(
)
2
2
2
0
2
x
yz
x
yz
x yz
−
≥ ⇔
+
≥
⇔
(
)
2
2
2
2
(
)(
)
(
)(
)
x
xy xz yz xy
x yz xz
x y x z
xy
xz
x y x z
xy
xz
+
+
+
≥
+
+
⇔
+
+ ≥
+
⇔
+
+
≥
+
munosabatni topamiz. Bundan
(
)(
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(
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(
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1
x
y
z
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x y x z
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x y z y
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z x z y
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y
yx
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x
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x
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+
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≤
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≤
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=
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+
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|