7-masala.
(XMO–1995
)
.
1
abc
=
shartni qanoatlantiruvchi ixtiyoriy
musbat sonlar uchun
, ,
a b c
3
3
3
1
1
1
2
(
)
(
)
(
)
a b c
b c a
c a b
3
+
+
≥
+
+
+
tengsizlikni isbotlang.
Yechilishi.
a
deb faraz qilamiz. U holda
b c
≥ ≥
1
1
1
c
b
a
≥ ≥
munosabatdan
1
1
1
ac bc
ab bc
ab ac
≥
≥
+
+
+
tengsizliklarga ega bo’lamiz.
21
Shuning uchun
1
1
1
(
)
(
)
(
c ac bc
b ab bc
a ab ac
≥
≥
+
+
)
+
.
ab ac bc
≥
≥
bo’lgani bois (
va
, , )
ab ac bc
1
1
1
,
,
(
) (
)
(
c ac bc b ab bc a ab ac
⎛
⎞
⎜
⎟
+
+
+
⎝
⎠
)
uchliklar bir xil tartiblangan bo’ladi.
(2) tengsizlikda
1
2
3
( , , )
a a a
=
1
1
1
,
,
(
) (
)
(
c ac bc b ab bc a ab ac
⎛
⎞
⎜
⎟
+
+
+
⎝
⎠
)
,
1
2
3
( , , )
b b b
=
,
( , , )
ab ac bc
1
2
3
( , , )
x x x
=
( , ,
)
ac bc ab
deb olsak,
(
)
(
)
(
)
(
)
(
)
(
ab
ac
bc
ac
bc
ab
c ac bc
b ab bc
a ab ac
c ac bc
b ab bc
a ab ac
+
+
≥
+
+
+
+
+
+
+
)
+
tengsizlikka,
1
2
3
( , , )
x x x
= (
deb olsak,
, , )
bc ab ac
(
)
(
)
(
)
(
)
(
)
(
ab
ac
bc
bc
ab
ac
c ac bc
b ab bc
a ab ac
c ac bc
b ab bc
a ab ac
+
+
≥
+
+
+
+
+
+
+
)
+
tengsizlikka ega bo’lamiz.
Ularni hadma-had qo’shib
1 1 1
2
(
)
(
)
(
)
ab
ac
bc
c ac bc
b ab bc
a ab ac
c b a
⎛
⎞
+
+
≥ + +
⎜
⎟
+
+
+
⎝
⎠
ga ega bo’lamiz.
shartni hisobga olib, o’rta qiymatlar haqidagi Koshi
tengsizligiga ko’ra
1
abc
=
3
1 1 1
1 1 1
3
3
c b a
c b a
+ + ≥
⋅ ⋅ =
.
Demak,
22
3
3
3
1
1
1
(
)
(
)
(
)
(
)
(
)
(
)
ab
ac
bc
c ac bc
b ab bc
a ab ac
a b c
b c a
c a b
+
+
=
+
+
+
+
+
+
+
+
3
2
≥
.
8-masala. (
XMO –1978
)
. {
a
1
,
a
2
, ...,
a
n
} – turli natural sonlardan iborat ketma-ketlik
bo’lsin.
2
1
1
1
n
n
k
k
k
a
k
k
=
=
≥
∑
∑
tengsizlik bajarilishini isbotlang.
Yechilishi.
(
i
1
,
i
2
, ...,
i
n
) – 1, 2, ...,
n
sonlarining shunday o’rin almashtirishi
bo’lsinki, ular uchun
bajarilsin.
1
2
...
n
i
i
a
a
a
<
< <
i
2
2
1
1
(
1)
1
n
n
2
1
<
<
<
−
…
bo’lgani uchun, (1) tengsizlikka ko’ra
2
2
1
1
k
n
n
i
k
k
k
a
a
k
k
=
=
≥
∑
∑
.
Ravshanki,
.
,
1, 2,...,
k
i
a
k k
≥
=
n
Bundan
2
2
2
1
1
1
1
k
n
n
n
n
i
k
k
k
k
k
a
a
k
k
k
k
=
=
=
=
≥
≥
=
∑
∑
∑
∑
1
k
0
.
9-masala.
(XMO-1964).
a
,
b
,
c
– biror uchburchakning tomonlari uzunliklari
bo’lsin.
2
2
2
(
)
(
)
(
) 3
a b c a
b c a b
c a b c
abc
+ −
+
+ − +
+ − ≤
tengsizlikni isbotlang.
Yechilishi.
a
≥
b
≥
c
deb faraz qilamiz. Dastlab quyidagini isbotlaymiz:
(
)
(
)
(
)
a b c a
b c a b
c a b c
+ −
≤
+ −
≤
+ −
.
Buning uchun
(
)
(
) (
)(
)
c a b c
b c a b
b c b c a
+ − −
+ −
=
−
+ −
≥
,
23
(
)
(
) (
)(
)
b c a b
a b c a
a b a b c
+ − −
+ −
=
−
+ − ≥
0
)
.
ekanligini eslatish kifoya.
Demak,
va
( , , )
a b c
(
)
(
), (
), (
a b c a b c a b c a b c
+ −
+ −
+ −
uchliklar turlicha
tartiblangan bo’ladi .
(3) tengsizlikda
1
2
3
( , , )
a a a
=
(
)
,
=
,
(
), (
), (
)
a b c a b c a b c a b c
+ −
+ −
+ −
1
2
3
( , , )
b b b
( , , )
a b c
1
2
3
( , , )
x x x
= (
deb olsak,
, , )
b c a
2
2
2
(
)
(
)
(
)
(
)
(
)
(
)
a b c a
b c a b
c a b c
ba b c a
cb c a b
ac a b c
+ − +
+ − +
+ − ≤
≤
+ − +
+ − +
+ −
tengsizlikka,
1
2
3
( , , )
x x x
= (
deb olsak,
, , )
c a b
2
2
2
(
)
(
)
(
)
(
)
(
)
(
)
a b c a
b c a b
c a b c
ca b c a
ab c a b
bc a b c
+ − +
+ − +
+ − ≤
≤
+ − +
+ − +
+ −
.
0
tengsizlikka ega bo’lamiz.
Ohirgi ikkita tengsizliklarni qo’shib va soddalashtirib, berilgan tengsizlikni hosil qilamiz.
10-masala.
(XMO-1983).
a
,
b
,
c
– biror uchburchakning tomonlari uzunliklari
bo’lsin.
2
2
2
(
)
(
)
(
)
a b a b
b c b c
c a c a
− +
− +
−
≥
tengsizlikni isbotlang.
Yechilishi.
Umumiylikka putur etkazmagan holda
a
≥
b
deb olamiz.
Agar
a
≥
b
≥
c
bo’lsa, u holda
1
1 1
a b c
≤ ≤
va oldingi masala yechimidan
(
)
(
)
(
)
c a b c
b c a b
a b c a
+ − ≥
+ − ≥
+ −
.
ga ega bo’lamiz.
24
Ya’ni
1 1 1
, ,
a b c
⎛
⎞
⎜
⎟
⎝
⎠
va
(
)
(
), (
), (
a b c a b c a b c a b c
)
+ −
+ −
+ −
uchliklar bir xil
tartiblangan.
(2) tengsizlikda
1
2
3
( , , )
a a a
=
(
)
,
=
(
), (
), (
)
a b c a b c a b c a b c
+ −
+ −
+ −
1
2
3
( , , )
b b b
1 1 1
, ,
a b c
⎛
⎞
⎜
⎟
⎝
⎠
,
1
2
3
( , , )
x x x
=
1 1 1
, ,
c a b
⎛
⎞
⎜
⎟
⎝
⎠
deb olsak,
1
1
1
(
)
(
)
(
)
a b c
a b c a
b c a b
c a b c
a
b
c
+ + =
+ −
+
+ − +
+ − ≥
1
1
1
(
)
(
)
(
)
a b c a
b c a b
c a b c
c
a
b
≥
+ − +
+ − +
+ −
tengsizlikni hosil qilamiz.
Soddalashtirishlardan so’ng bu tengsizlik berilgan tengsizlikka tengkuchli bo’lgan
ushbu
1
1
1
(
)
(
)
(
)
a b a
b c b
c a c
c
a
b
0
−
+
− +
− ≤
tengsizlikka keladi.
a
≥
c
≥
b
holni tahlil qilishni o’quvchilarga qoldiramiz.
11-masala.
(4-Xalqaro Jautikov olimpiadasi, Almati, 2008 yil)
1
abc
=
shartni qanoatlantiruvchi ixtiyoriy
musbat sonlar uchun
, ,
a b c
1
1
1
(
)
(
)
(
)
a b b
b c c
c a a
3
2
+
+
≥
+
+
+
tengsizlikni isbotlang.
Yechilishi.
Tengsizlikning chap tomonini
S
orqali belgilaymiz.
25
a b c
≥ ≥
deb faraz qilamiz. U holda
1
1 1
a b c
≤ ≤
va
1
1
1
b c
c a
a b
≥
≥
+
+
+
tengsizliklar
o’rinli, ya’ni
1 1 1
, ,
a b c
⎛
⎞
⎜
⎟
⎝
⎠
va
1
1
1
,
,
b c c a a b
⎛
⎜
+
+
+
⎝
⎠
⎞
⎟
uchliklar turlicha tartiblangan bo’ladi.
(3) tengsizlikda
1
2
3
( , , )
a a a
=
1
1
1
,
,
b c c a a b
⎛
⎞
⎜
⎟
+
+
+
⎝
⎠
,
=
1
2
3
( , , )
b b b
1 1 1
, ,
a b c
⎛
⎞
⎜
⎟
⎝
⎠
,
1
2
3
( , , )
x x x
=
1 1 1
, ,
c a b
⎛
⎞
⎜
⎟
⎝
⎠
deb olsak
1
1
1
1
1
1
(
)
(
)
(
)
(
)
(
)
(
)
T
S
a b c
b c a
c a b
c b c
a c a
b a b
=
+
+
≤
+
+
+
+
+
+
+
+
=
tengsizlikga ega bo’lamiz.
O’rta qiymat haqidagi Koshi tengsizligini va
1
abc
=
shartni hisobga olib,
quyidagilarga ega bo’lamiz:
3
1
1
1
1
1
1
2
(
)
(
)
(
)
(
)
(
)
(
)
3
3
(
)
(
)
(
)
(
)
(
)
(
)
S S T
a b b
a b c
b c c
b c a
c a a
c a b
b c
c a
a b
b c
c a
a b
a b bc
b c ca
c a ab
a b bc b c ca c a ab
⎛
⎞ ⎛
⎞⎛
≥ + =
+
+
+
+
=
⎜
⎟ ⎜
⎟⎜
+
+
+
+
+
+
⎝
⎠ ⎝
⎠⎝
+
+
+
+
+
+
=
+
+
≥
⋅
⋅
=
+
+
+
+
+
+
⎞
⎟
⎠
Bundan
1
1
1
(
)
(
)
(
)
a b b
b c c
c a a
+
+
+
+
+
3
2
≥
kelib chiqadi.
26
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