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
...
)