Sonli tengsizliklar haqida. Toshkent 2008 - bet 25

41
1
2
2
2
2
2
2
2
1
1
2
1
2
2
2
2
1
2
2
2
2
2
2
1
1
2
1
....
1
1
1
...
...
1
1
1
...
n
n
n
n
x
x
x
x
x
x
x
x
x
x
x
x
n
x
x
x
x
x
+
+
+
<
+
+
+
+
+
+ +
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
⎜
⎟
<
+
+ + ⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
+
+
+
+
+ +
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
munosabatni hosil qilamiz. Bundan
2
2
2
1
2
2 2
2
2 2
2
2
2 2
1
1
2
1
2
..
1
(1
)
(1
)
(1
...
)
n
n
x
x
x
x
x
x
x
x
x
+
+ +
<
+
+
+
+
+
+ +
ekanligini ko’rsatsak, yuqoridagi tengsizlik isbotlanadi.
(
)
2
2
2
2
2
2
2
2
2
2
1
2
1
1
1
2
2
2
2
2
1
1
1
(1
...
)(1
...
)
1
...
1
1
1
...
1
...
k
k
k
k
k
k
k
x
x
x
x
x
x
x
x
x
x
x
x
x
x
−
−
≤
=
+
+
+ +
+
+ +
+
+
+ +
=
−
+
+ +
+
+ +
bo’lgani
uchun
2
2
2
2
2
1
1
1
1
1
1
1
...
1
...
n
k
k
k
n
x
x
x
x
x
=
⎛
⎞
< −
<
⎜
⎟
+
+ +
+
+ +
⎝
⎠
∑
.
53.
3
3
3
3
9(
) (
)
a
b
c
a b c
+
+
≥
+ +
tengsizlikdan foydalanib,
( )
(
)
3
3
3
3
2
2
2
3
3
3
3
2
3
3
3
1
1
1
1
1
1
6
6
6
9
6
6
6
4 3
( )
( )
3
1
1
1
1
3
3
3
3
3
3
(
)
4 3
3
3
3
3
b
c
a
b
c
a
a
b
c
a
b
c
ab
bc
ca
ab
bc
ac
abc
c
a
b
abc
ab bc ca
abc
abc
⎛
⎞
+
+
+
+
+
≤
+
+ +
+ +
=
⎜
⎟
⎝
⎠
−
+
+
−
−
−
=
+
+
+
=
≤
+
+
−
≤
=
munosabatni hosil qilamiz.Endi
3
3
1
abc
abc
≤
yoki
2 2 2
1
27
a b c
≤
ekanligi
ko’rsatamiz:
3
2
1
(
)
( )( )( )
3
27
ab bc ca
abc
ab bc ca
+
+
⎛
⎞
=
≤
=
⎜
⎟
⎝
⎠
.
42
54.
1
(
)
bc a b c
−
=
+
ekanligidan
2
2
3
2
2
3
3
3
2
3
3
3
3
3
(
)
(1
)
2
(
)
2
1
1
1
1
3
3
3
3
2
3
9
abcx
a b c c b x
a b c c b x
ax
bc
bc
bc
bc
ax
bc
bc
bc ax
bc
ax
bc
=
⋅
⋅
≤
+
+
=
+
−
=
⎛
⎞
+
−
⎜
⎟
⎛
⎞
=
+
−
+
≤
+
+
=
+
+
⎜
⎟
⎜
⎟
⎝
⎠
⎜
⎟
⎝
⎠
munosabatni hosil qilamiz. Xuddi shunday
3
3
1
1
1
1
3
, 3
3
9
3
9
abcy by
ac
abcz cz
ab
≤
+
+
≤
+
+
tengsizliklar o’rinli.bu tengsizliklarni hadma-had qo’shib
isboti talab etilgan
tengsizlikni hosil qilamiz.
55.
Tengsizlikni ikkala qismiga umumiy maxraj tanlab,
1
1
1
1
1
1
1
1
1
1
1
n
n
n
n
i
i
i
i
i
i
i
i
a
a
n
a
a
=
=
=
=
−
≥
⋅
+
+
∑
∑
∑ ∑
yoki
1
1
1
1
1
1
(
1)
1
n
n
n
i
i
i
i
i
i
i
n
a a
a
a
=
=
=
≥
+
+
∑
∑
∑
munosabatni hosil qilamiz. Umumiylikni chegaralamasdan
1
2
...
n
a
a
a
≥
≥ ≥
deb
olsak, oxirgi tengsizlik Chebishev tengsizligia ko’ra o’rinlidir.
56.
1
2
1
....
n
a
a
a
k
−
+
+
+
=
deb olamiz. U holda isboti talab
etilgan tengsizlikka teng
kuchli bo’lgan
0
0
2
2
2
0
0
0
0
1
1
(
) (
1)(
)(
)
(
)
(
1).
n
n
n
n
n
n
a
k a
k
a
k a
k
n
n
n
n
n k a
a
k
n
a
k a
k
k k a
a
a a n
+ +
+
+
⋅
≥
⋅
⇔
+
−
+
+
≥
−
+
+
⇔
+
+
≥
−
tengsizlikni hosil qilamiz.
0
0
2
n
n
a
a
a a
+
≥
ekanligidan
(
)
0
1
n
k
n
a a
≥
−
tengsizlikni isbotlasak yuqoridagi tengsizlik isbotlandi.
2
1
1
(
1,2,....,
1)
i
i
i
a
a
a
i
n
−
+
⋅
≤
=
−
tengsizlikka ko’ra
0
1
2
1
1
2
3
...
n
n
a
a
a
a
a
a
a
a
−
≤
≤
≤ ≤
yoki
0
1
1
2
2
......
n
n
n
a a
a a
a a
−
−
≤
≤
≤
munosabatni hosil qilamiz. Bundan,
43
1
2
1
1
1
2
2
1
1
2
2
0
0
0
0
...
(
) (
) ... 2
1
2
... 2
2
... 2
(
1)
2
n
n
n
n
n
n
n
n
n
K a
a
a
a
a
a
a
a a
n
a a
a a
a a
a a
n
a a
−
−
−
−
−
= +
+ +
=
+
+
+
+ ≥
+
−
+
+ ≥
+
+ = ⋅
=
−
ekanligi kelib chiqadi.
57.
Umumiylikni chegaralamasdan
a b c
≤ ≤
deb olamiz. U holda o’rta arifmetik
va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini quyidagi usulda qo’llab,
[
]
2
8
8 (1
) 2(
) (1
) 2(
) (
)(1
)
2
a b
abc
ab
a b
a b
a b
a b
a b
a b
+
=
− − ≤
+
− −
=
+
+
− −
≤
tengsizlikni topamiz.
1
1
(
)(
) (
)(
) 0
2
2
a c x
b c y
−
−
+ −
−
≥
munosabat o’rinli
ekanligidan
8
2
a b
ax by cz
abc
+
+
+
≥
≥
tengsizlikni to’g’riligini topamiz.
58.
1
1
2
1
...
n
n
x
x
x
x
+
= − −
− −
bo’lsin.
U holda
1
0
n
x
+
>
va o’rta arifmetik va o’rta
geometrik miqdorlar haqidagi Koshi tengsizligini quyidagi usulda qo’llasak
1 2
1
1
2
1
...
1
...
(
1,2,...,
1)
n
n
i
n
i
i
x x x
x
x
x
x
x
n
i
n
x
+
+
− = +
+ +
− ≥
=
+
munosabatni olamiz.
Bu tengsizliklarni hadma-had ko’paytirib,
1
1
1
1
1 2
1
1 2
1
1 2
1
2
1
1
...
(1
)
...
... (1
...
)