Sonli tengsizliklar haqida. Toshkent 2008 - bet 24

36
37.
Bu tengsizlikni shakl almashtirish natijasida
1
1
1
0
(
1)
(
1)
(
1)
y
z
x
z
x
y
x
y
y
z
z
x
⎛
⎞ ⎛
⎞ ⎛
⎞
+
+
+
−
+
−
+
−
≥
⎜
⎟ ⎜
⎟ ⎜
⎟
+
+
+
⎝
⎠ ⎝
⎠ ⎝
⎠
yoki
3
1
1
1
x z
x y
y z
x
y
z
+
+
+
+
+
≥
+
+
+
(*)ga teng kuchli tengsizlikka olib kelamiz.
Koshi-Bunyakovskiy-Shvarts tengsizligiga ko’ra
(
)(
)
1
x
x xy x z
+ ≤
+
+
munosabat o’rinli. Bundan
2
(
1)
(1
)
x
x z
x
y
+
+ ≥
+
yoki
1
1
(1
)
x z
x
x
x
y
+
+
≥
+
+
munosabatni
hosil qilamiz.
Demak
3
1
1
1
1
1
1
(1
)
(1
)
(1
)
1
1
1
3
3
(1
)
(1
)
(1
)
x z
x y
z y
x
y
z
x
y
z
x
y
y
z
z
x
x
y
z
x
y
y
z
z
x
+
+
+
+
+
+
+
+
≥
+
+
≥
+
+
+
+
+
+
+
+
+
≥
⋅
⋅
=
+
+
+
38.
Avval
0
t
∀ >
uchun
2
3
6
3
3(
1)
1 (*)
t
t
t
t
− +
≥ + +
tengsizlik o’rinli ekanligini
ko’rsatamiz. (*) ni shakl almashtirib, quyidagi munosabatni hosil qilamiz:
4
2
(
1) (2
2) 0
t
t
t
−
− +
≥
bu
tengsizlik
0
t
∀ >
uchun o’rinli. Bundan
2
3
2
3
2
3
6
3
6
3
6
3
3(
1) 3(
1) 3(
1)
(
1)(
1)(
1) (**)
x
x
y
y
z
z
x
x
y
y
z
z
− +
− +
− +
≥
+
+
+
+
+
+
munosabatni hosil qilamiz. Umumlashgan Koshi-Bunyakovskiy-Shvarts
tengsizligini qo’llasak,
6
3
6
3
6
3
2 2 2
3
(
1)(
1)(
1) (
1) (***)
x
x
y
y
z
z
x y z
xyz
+
+
+
+
+
+ ≥
+
+
(**) va (***) larni hadma-had ko’paytirib, isboti talab
etilgan tengsizlikni hosil
qilamiz.
39.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini
qo’llaymiz:
37
1
1
1
1
1
2 3
1
1 2 3
1
1
1
1
...
1
1
...
1
...
2
3
2 3 4
2 3 4
1
.
n
n
n
n
n
n
n
n
n
n
n
n
n
−
−
−
⎛
⎞ ⎛
⎞
⎛
⎞
+ −
+ −
+ + −
= + + + + +
≥
⋅ ⋅ ⋅
=
⎜
⎟ ⎜
⎟
⎜
⎟
⎝
⎠ ⎝
⎠
⎝
⎠
=
=
40.
Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llab,
1
1
1
1
1
1
x
y
z
x
y
z
x y z
x y z
x
y
z
−
−
−
− +
− +
− ≤
+ + ⋅
+
+
=
+ +
munosabatni hosil qilamiz.
41.
Berilgan tengsizlikdan
3
3
3
2 2 2
2
2
2 2
(
)(
) 9
(
)
a
b
c a b c
a b c
a
b
c
+
+
+ + ≥
=
+
+
tengsizlikni hosil qilamiz. Bu tengsizlik esa Koshi -
Bunyakovskiy tengsizligiga
ko’ra o’rinlidir.
42.
O’rta arifmetik va o’rta geometrik miqdorlar o’rtasidagi munosabatni quyidagi
usulda qo’llaymiz:
2
2
2
2
2
2
2
2
2
2
2
3
4
2
2 2 2
2 2 2
1
1
1
1
1
9
1
1
1
8 1
1
1
(
)
9 (
)
9
4
9
8
1
4
3
8
3
9 (
)
3
9 (
)
3
(
)
28 3
9
(
)
x y t
x
y
z
x y z
x
y
z
x
y
z
x y z x y z
x y z
x y z xyz
xyz x y z
xyz x y z
⎛
⎞
⎛
⎞
+
+
+
=
+
+
+
+
+
+
≥
⎜
⎟
⎜
⎟
+ +
+ +
⎝
⎠
⎝
⎠
≥
+
≥
+
=
+ +
+ +
+ +
=
+ +
43.
k
k
d
n a
= −
belgilash kiritsak,
u holda
(
)
2
2
2
2
3
2
3
1
1
1
1
2
2
1 1
n
n
n
n
k
k
k
k
k
k
k
k
d
n a
n n
n
a
a
n
n n
n
=
=
=
=
=
−
= ⋅
−
+
≤
−
⋅
+
+ =
∑
∑
∑ ∑
munosabatni hosil bo’ladi. Bundan
1
k
d
≤
yoki
1
1
k
n
a
n
− ≤
≤ +
.
38
44.
Tengsizlikni ikkala qismini
a b c
+ +
ga bo’lib va
,
,
a
b
c
x
y
z
a b c
a b c
a b c
=
=
=
+ +
+ +
+ +
belgilash
kiritsak
(
)(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1 1
1.
2
2
2 2
x
y
z
x
y
z
xy yz zx
xy yz zx
x y z
α
β
γ
α
β
γ
αβ βγ γα
γα αβ βγ
α β γ
+
+
+
+
+
+
+
≤
+
+
+
+
+
+
+
+
+
+
+
+
=
+ +
+
+ +
= + =
45.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini
quyidagi usulda qo’llaymiz:
3
2
2
3
3
3
3 3
3
3
1
1
1
1
1
1
1
1 1
1
2
1
8
(
)
2
3
3
3
3
3 ( )
64
3(
)
a
ab
a b b
a
b
ab a b
a b
ab ab
ab
a b
⎛
⎞
+
+
+
=
+
+
+
≥
+
=
≥
⎜
⎟
⎝
⎠
≥
+
46.
[ ]
1
2
6
, ....
0;1
x x
x
∈
ekanligidan tengsizlikning chap qismi quyidagi ifodadan
kichik
yoki teng
3
3
3
1
2
6
5
5
5
5
5
5
5
5
5
1
2
6
1
2
6
1
2
6
3
3
3
1
2
6
5
5
5
1
2
6
...
.....
4
......
4
.....
4
...
3
.(*)
...
4 5
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
+
+ +
=
+
+
+
+
+
+
+
+
+
+
+
+
+
+ +
=
≤
+
+ +
+
Ixtiyoriy
0
t
≥
uchun
(
)
(
)
2
5
3
3
2
3
2 5
1
3
6
4
2
0
t
t
t
t
t
t
+ ≥
⇔ −
+
+ +
≥
munosabat
o’rinli ekanligidan foydalansak (*) kelib chiqadi.
47.
Tengsizlikni chap qismini
S
bilan belgilab,
Koshi-Bunyakovskiy-Shvarts
tengsizligini quyidagi usulda qo’llab,
(
)
(
)
2
2
2
2
2
2 2
1 2
(1 2 )
(1 2 )
(
)
S a
bc
b
ac
c
ab
a
b
c
⋅
+
+
+
+
+
≥
+
+
yoki
1
1 2
(
)
S
abc a b c
≥
+
+ +
39
munosabatni hosil qilamiz.
2
2
2
3
(
)
a
b
c
abc a b c
+
+
≥
+ +
tengsizlikka ko’ra
(
)
2
2
2
2
1
1
3
2
2
5
1
3
(
) 1
3
3
S
abc a b c
a
b
c
≥
≥
=
+
+ +
+
+
+
ekanligini hosil qilamiz.
48.
Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llab,
(
)
(
)
(
)
2
2
2
2
2
2
3
3
3
2
3
1
1
3
3
3
2
1
2
3
1
1
2
1
(
1)
...
1 2
3
... (
1)
2
2
3
(
1)
2
...
, 2
2001
...
1
i
i
i
i
i
i i
x
x
x
x
x
i
i
x
x
x
x
x
ёки
i
x
x
x
i i
−
−
−
⎛
⎞
−
⎛
⎞ ⋅ ≥
+
+
+ +
+
+ + + −
≥
⎜
⎟
⎜
⎟
−
⎝
⎠
⎝
⎠
≥
+
+ + +
≥
≤ ≤
+
+ +
−
ekanligini topamiz. Bundan
2001
2
1
2
1
1
1
1
1
2
...
2 1
1,999.
...
1 2 2 3
2000 2001
2001
i
i
i
x
x
x
x
=
−
⎛
⎞
⎛
⎞
≥
+
+ +
=
−
>
⎜
⎟
⎜
⎟
+
+ +
⋅
⋅
⋅
⎝
⎠
⎝
⎠
∑
49.
Tengsizlikni
1
1
1
3
a c a
b a b
c
b c
b
c b
c
a c
a b a
+
+
+
⎛
⎞ ⎛
⎞ ⎛
⎞
−
+ +
−
+ +
−
+ ≥
⎜
⎟ ⎜
⎟ ⎜
⎟
+
+
+
⎝
⎠ ⎝
⎠ ⎝
⎠
yoki
2
2
2
3
(
)
(
)
(
)
b
ac
c
ab
a
bc
b c b
c a c
a b a
+
+
+
+
+
≥
+
+
+
kurinishda yozamiz.Koshi-Bunyakovskiy-
Shvarts tengsizligini quyidagi usulda qo’llasak,
(
)
(
)
2
2
2
3
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
3
3.
(
)
(
)
(
)
b c a
c a b
b
ac
c
ab
a
bc
b a c
b c b
c a c
a a b
b a c
c a b
a b c
b c a b a c c a b
b a c c a b a b c
+
+
+
+
+
+
+
+
≥
+
+
≥
+
+
+
+
+
+
+
+
+
≥
⋅
⋅
=
+
+
+
50.
Avval tengsizlikni induktsiya metodi yordamida
2 ,
k
n
k N
=
∈
sonlar uchun
isbotlaymiz. 2
n
=
da
(
)(
)
2
1 2
1
2
1
2
1 2
1
2
1 2
1
1
1
2
0
1
1
1 (
1)(
1)(
1)
y y
y
y
y
y
y y
y
y
y y
−
−
+
−
=
≥
+
+
+
+
+
+
2
k
n
=
da tengsizlik o’rinli bo’lsin deb
faraz qilsak, u holda
1
2
k
n
+
=
da
40
1
1
1
1
1
2
2
2 1
2
2
2
1
2
2
2
1 2
1
2
2
2 1 2
2
2
2
1
1
1
1
1
1
...
....
1
1
1
1
1
1
2
2
2
1
...
1
...
1
....
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
y
y
y
y
y
y
y y
y
y
y
y
y y
y
+
+
+
+
+
+
+
+
+
⎛
⎞ ⎛
⎞
+
+ +
+
+
+
+
≥
⎜
⎟ ⎜
⎟
⎜
⎟ ⎜
⎟
+
+
+
+
+
+
⎝
⎠ ⎝
⎠
≥
+
≥
+
+
+
⋅
munosabat o’rinli va tengsizlik
2 ,
,