Sonli tengsizliklar haqida. Toshkent 2008 - bet 22

28
2 (
)
2 (
)
2 (
)
(
)(
) (
)(
) (
)(
)
2(
) 2(
) 2(
)
a
b
c
a a b c
b a b c
c a b c
a b
b c
c a
a b a c
b c b a
c a c b
a c
b a
c b
a b c
a b c
a b c
⎛
⎞
+ +
+ +
+ +
+
+
≤
+
+
×
⎜
⎟
+
+
+
+
+
+
+
+
+
⎝
⎠
⎛
⎞
+
+
+
×
+
+
=
⎜
⎟
+ +
+ +
+ +
⎝
⎠
2 (
)
(
)(
) (
)(
) (
)(
)
a
b
c
a b c
a b a c
b c b a
c a c b
⎛
⎞
=
⋅ + +
+
+
⎜
⎟
+
+
+
+
+
+
⎝
⎠
munosabatni hosil qilamiz. Endi
(
)
(
)(
) (
)(
) (
)(
)
2(
)(
)
9
(
)(
)(
)
4
a
b
c
a b c
a b a c
b c b a
c a c b
a b c ab ac bc
a b b c c a
⎛
⎞
+ +
+
+
=
⎜
⎟
+
+
+
+
+
+
⎝
⎠
+ +
+
+
=
≤
+
+
+
yoki
(
)
8
(
) 9(
)(
)(
)
6
(
)
(
)
(
)
a b c ab bc ca
a b b c c a
abc ab a b
bc b c
ac a c
+ +
+
+
≤
+
+
+
⇔
≤
+ +
+ +
+
tengsizlikni isbotlash yetarli. Bu tengsizlik esa o’rta arifmetik va o’rta
geometrik
miqdorlar o’rtasidagi munosabatga ko’ra o’rinli. Bulardan yuqoridagi isboti talab
etilgan tengsizlik isbotlandi.
22.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan
quyidagicha foydalanamiz:
19
17
11
3
4
2
5
3
4
20
15
12
2
1
1
1
1
15
20
12
12
20
1
1
1
1
20
20
12
12
(
3 )(
4 )(
2 ) (
)(
)(
)
4
5
3
60
60
60
60
60
a
b b
c c
a
a b b b b c c c c c a a
ab
b c
a c
a
b
c
c
c
c
c
c
abc
abc
abc
abc
a
b
a
b
a b
+
+
+
=
+ + +
+ + + +
+ +
≥
≥
⋅
⋅ ⋅
=
⋅
⋅
=
⋅
⎛ ⎞
⎛ ⎞
=
=
=
⋅
≥
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠
⋅
23.
,
,
x
y
t
a
b
c
y
t
x
=
=
=
deb
belgilash kiritsak,
u holda
29
2
3
3
3
3
2
2
2
4
3 2
xt
yx
yt
x
y
t
y
t
x
y
t
x
y
t
x
x
y
t
⎛
⎞
⎛
⎞
+
+
≤
+ + + + + +
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
bundan
2
3
3
1
1 1
4
3 2
,
x
y
t
y
t
x
xyt
y
x t
y
t
x
t
y
t
⎛
⎞
⎛
⎞
+ +
≤
+ + + + + +
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
2
3
2
3
4
(
) 3 2
(
)
x t
t y t x
xy yt tx
y
x
t
xyt
⎛
⎞
+
+
+
+
+
≤
+
+
+
⎜
⎟
⎝
⎠
.
Bundan
3
2
64(
)
27((
)(
)
)
xy yt tx
xy yt tx x y t
xyt
+
+
≤
+
+
+ + −
tengsizlikni
isbotlasak yetarli.
(
)
2
2
2
2
2
3
27((
)(
)
)
(
)(
)
27 (
)(
)
9
8
27
(
)(
)
64(
)
64(
)
9
3
x y t xy yt tx
xyt
x y t xy yt tx
x y t xy yt tx
x y t
x y t xy yt tx
xy yt tx
xy yt tx
+ +
+
+
−
≥
+ +
+
+
⎛
⎞
≥
+ +
+
+
−
=
⎜
⎟
⎝
⎠
+ +
⎛
⎞
=
+ +
+
+
=
+
+
≥
+
+
⎜
⎟
⎝
⎠
24.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligiga ko’ra
1
3
3
1 1 1
1 1 1
3
,
3
a b c
abc
a b c
a b c
⎛
⎞
+ + ≥
⋅ ⋅
+ + ≥
⎜
⎟
⎝
⎠
tengsizliklar o’rinli. Bulardan
1 1 1
(
)(
) 9
a b c
a b c
+ +
+ +
≥
ekanligini topamiz. Bu
tengsizlik va
2
2
2
3(
)
a b c
a
b
c
+ + ≤
+
+
tengsizliklarni hadma-had ko’paytirib,
(
)
2
2
2
1 1 1
3 3
a
b
c
a b c
⎛
⎞
≤
+ +
+
+
⎜
⎟
⎝
⎠
va bundan
2
2
2
2
2
2
2
2
2
2
2
2
3 1 1 1 1
(
)
3(
)
3 3
(
)
.
a
b
c
a
b
c
a
b
c
a b c
a b c
a
b
c
+ ⎛
⎞
+ +
+
+
≥
+
+
+
+
+
≥
⎜
⎟
⎝
⎠
≥
+ + +
+
+
25.
Musbat
x
son
uchun
2
1
x
−
va
3
1
x
−
ifodalar bir xil ishoraladir, ya’ni
30
2
3
5
3
2
0 (
1)(
1)
1
x
x
x
x
x
≤
−
− =
−
−
+
yoki
5
2
3
3
2
x
x
x
−
+ ≥
+
.Bu tengsizlikdan
foydalansak, u holda
5
2
5
2
5
2
3
3
3
(
3)(
3)(
3) (
2)(
2)(
2)
a
a
b
b
c
c
a
b
c
−
+
−
+
−
+ ≥
+
+
+
.
Bundan
(
)
3
3
3
3
2 (
2)(
2) (
)
a
b
c
a b c
+
+
+
≥
+ +
tengsizlikni isbotlash yetarli.
Umumlashgan Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda
qo’llaymiz:
3
3
3
3
3
3
3
(
2)(
2)(
2) (
1 1)(1
1)(1 1
) (
)
a
b
c
a
b
c
a b c
+
+
+
=
+ +
+
+
+ +
≥
+ +
26.
Umumlashgan Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda
qo’llaymiz:
2
2
2
2
2
2
2
2
2
2
2
2
3
(
)(
)(
)
(
)(
)(
) (
)
a
ab b b
bc c c
ca a
a
ab b c
b
bc ac a
c
ac ab bc
+
+
+
+
+
+
=
=
+
+
+
+
+
+
≥
+
+
27.
Agar , ,
a b c
>1 bo’lsa,
2
2
2
4
a
b
c
abc
+
+
+
>
bo’ladi. Agar
1
a
≤
bo’lsa, u
holda (1
) 0
ab bc ca abc bc abc bc
a
+
+
−
≥
−
=
−
≥
.Endi 2
ab bc ca abc
+
+
−
≤
tengsizlikni isbotlaymiz.
2cos ,
2cos ,
2cos
a
A b
B c
C
=
=
=
va , ,
0,
2
A B C
π
⎡
⎤
∈ ⎢
⎥
⎣
⎦
deb belgilash kiritsak, shartga ko’ra
A B C
π
+ + =
ekanligini
topamiz va
1
cos cos
cos cos
cos cos
2cos cos cos
2
A
B
B
C
A
C
A
B
C
+
+
−
≤
tengsizlikni isbotlasak yetarli bo’ladi. Faraz etaylik,
3
A
π
≥
yoki 1 2cos
0
A
−
≥
.Bundan
cos cos
cos cos
cos cos
2cos cos cos
cos (cos
cos ) cos cos (1 2cos )
A
B
B
C
C
A
A
B
C
A
B
C
B
C
A
+
+
−
=
=
+
+
−
Quyidagi
3
cos
cos
cos
2
B
C
A
+
≤ −
va
2cos cos
cos(
) cos(
) 1 cos
B
C
B C
B C
A
=
−
+
+
≤ −
tengsizliklardan foydalansak,
(
)
3
1 cos
cos (cos
cos ) cos cos (1 2cos ) cos
cos
1 2cos
2
2
A
A
B
C
B
C
A
A
A
A
−
⎛
⎞ ⎛
⎞
+
+
−
≤
−
+
−
⎜
⎟ ⎜
⎟
⎝
⎠ ⎝
⎠
31
28.
0
x
≠
bo’lgani uchun
2
2
0
x
y
+
>
bo’ladi. Bundan
(
)
(
)
(
)
2
2
2
2
2
2
2 2
2
2
2
1
(
)
x
y
xy
x
y
x
y
−
+
=
+
+
ekanligini topamiz. Oxirgi tenglikda har bir qo’shiluvchi [-1;1] oraliqqa tegishli
ekanligidan
(
)
(
)
2
2
2
2
2
2
sin
xy
x
y
α
=
+
va
(
)
(
)
2
2
2
2
2
2
2
cos
x
y
x
y
α
−
=
+
deb belgilash kiritish mumkin. Bundan
(
)
(
)
(
)
(
)
2
2
2
2
2
4
2
2
2
2
2
4
4
cos
sin
xy x
y
x
y
x
y
α
α
−
=
=
+
+
,
2
2 2
2
16
(
)
16
sin 2
x
y
α
+
=
≥
,
2
2
4
x
y
+
≥
.
29.
4
3
,
,
3
5
2
x
y
z
a
b
c
=
=
=
deb belgilash kiritsak, u
holda masalaning sharti
quyidagicha ko’rinishga ega bo’ladi: 7
3
5
15
xy
yz
xz
+
+
≤
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini
yuqoridagi tengsizlikka qo’llab
12 10 8
15
15 7
3
5
15
xy
yz
xz
x y z
≥
+
+
≥
yoki
6 5 4
1
x y z
≤
(*) tengsizlikni topamiz.
Endi (*) dan foydalansak:
6
5
15
6 5 4
4
1
2 3
3
5
2
1
1
1
1
( , , )
...
...
2
2
2
2
2
1
1
15
1
15
...
2
2
2
2
та
та
та
P a b c
a b c
x
y
z
x
x
y
y
z
z
x y z
= + + = +
+ =
+ +
+
+ +
+
+
+ +
≥
≥
Tenglik 1
x y z
= = =
yoki
1
4
2
,
,
3
5
3
a
b
c
=
=
=
bo’lganda bajariladi.