Sonli tengsizliklar haqida. Toshkent 2008 - bet 20

72.
(Belorussiya -2002) Natural ,
a b
sonlar uchun
(
)
1
2
2
a
b
a b
− >
+
tengsizikni
isbotlang.
15
73.
(Belorussiya -2002) Musbat haqiqiy
, , ,
a b c d
sonlar uchun
(
) (
)
(
) (
)
(
) (
)
2
2
2
2
2
2
2
2
2
2
2
ad bc
a c
b d
a
b
c
d
a c
b d
a c
b d
−
+
+ +
≤
+ +
+
≤
+
+ +
+
+
+ +
tengsizlikni isbotlang.
74.
(APMO -2002) Musbat , ,
a b c
sonlar
1 1 1
1
a b c
+ + =
shartlarni qanoatlantirsa,
a bc
b ac
c ab
abc
a
b
c
+
+
+
+
+
≥
+
+
+
ttengsizlikni isbotlang.
75.
(APMO -1996) , ,
a b c
uchburchak tomonlari bo’lsa,
a b c
b c a
c a b
a
b
c
+ − +
+ − +
+ − ≤
+
+
tengsizlikni isbotlang.
76.
Musbat
, ,
a b c
sonlar
3
a b c
+ + =
shartni qanoatlantirsa,
3
3
3
2
2
2
2(
) 3(
1)
a b b c c a
a b b c c a
+
+
≥
+
+
−
tengsizlikni isbotlang.
77.
(APMO -1998) , ,
a b c
musbat sonlar uchun
3
1
1
1
2 1
a
b
c
a b c
b
c
a
abc
+ +
⎛
⎞
⎛
⎞⎛
⎞⎛
⎞
+
+
+
≥
+
⎜
⎟⎜
⎟⎜
⎟
⎜
⎟
⎝
⎠⎝
⎠⎝
⎠
⎝
⎠
tengsizlikni isbotlang.
78.
(Singapur -2001)
n N
∈
va
1
2
, ,...,
n
a a
a
sonlar
1
1
n
i
i
a
=
=
∑
shartni qanoatlantirsa,
4
4
4
1
2
2
2
2
2
2
2
1
2
2
3
1
1
...
2
n
n
a
a
a
a
a
a
a
a
a
n
+
+ +
≥
+
+
+
tengsizlikni isbotlang.
79.
(XMO -2000) Musbat , ,
a b c
sonlar
1
abc
=
shartni qanoatlantirsa,
1
1
1
1
1
1
1
a
b
c
b
c
a
⎛
⎞⎛
⎞⎛
⎞
− +
− +
− +
≤
⎜
⎟⎜
⎟⎜
⎟
⎝
⎠⎝
⎠⎝
⎠
tengsizlikni isbotlang.
16
80.
(Qozog’iston -2000) Yig’indisi birga teng bo’lgan
a, b, c
sonlar uchun
7
7
7
7
7
7
5
5
5
5
5
5
1
3
a
b
b
c
c
a
a
b
b
c
c
a
+
+
+
+
+
≥
+
+
+
tengsizlikni isbotlang.
81.
(Yaponiya -2002) Musbat ,
x y
sonlar uchun
2
1
7
2
2
x y
x y
xy
+ +
+
≥
+
tengsizlikni isbotlang.
82.
(Pol’sha -1996) Musbat , ,
a b c
sonlarning yig’indisi birga teng bo’lsa,
2
2
2
9
1
1
1 10
a
b
c
a
b
c
+
+
≤
+
+
+
tengsizlik o’rinli bo’lishini isbotlang.
83.
(Gonkong -2005) Musbat , , ,
a b c d
sonlarning yig’indisi birga teng bo’lsa,
(
) (
)
3
3
3
3
2
2
2
2
1
6
8
a
b
c
d
a
b
c
d
+
+ +
≥
+
+
+
+
tengsizlikni isbotlang.
84.
(Kanada -1998) Istalgan natural
n
soni
(
)
2
n
≥
uchun
1
1 1
1
1 1 1 1
1
1
...
...
1
3 5
2
1
2 4 6
2
n
n
n
n
⎛
⎞
⎛
⎞
+ + + +
>
+ + + +
⎜
⎟
⎜
⎟
+
−
⎝
⎠
⎝
⎠
tengsizlikni isbotlang.
85.
(Bosniya -2002) Agar
0
a
>
va 0
1
b
< <
bo’lsa,
2
2
1
1
a
b
a
b
b
a
+
+
−
≤ +
tengsizlikni isbotlang.
86.
(Polsha -1995) Agar
1
1
2
x
=
va
1
2
3
,
1
2
n
n
n
x
x
n
n
−
−
=
>
bo’lsa,
1
1
n
i
i
x
=
<
∑
tengsizlik
o’rinli bo’lishini isbotlang.
17
87.
(Bosniya -2002) Agar
0;
(
1,2,..., )
2
i
x
i
n
π
⎛
⎞
∈
=
⎜
⎟
⎝
⎠
sonlari
1
n
i
i
tgx
n
=
≤
∑
shartni
qanoatlantirsa,
2
1
2
sin
sin
... sin
2
n
n
x
x
x
−
⋅
⋅ ⋅
≤
tengsizlikni isbotlang.
88.
(Belorussiya -2000) Musbat , , ; , ,
a b c x y z
sonlar uchun
(
)
(
)
2
2
2
6
6
6
3
a
b
c
a
b
c
x
y
z
x y z
+
+
+
+
≥
+ +
tengsizlikni isbotlang.
89.
(AQSh -1997) Istalgan musbat
, ,
a b c
sonlar uchun
3
3
3
3
3
3
1
1
1
1
a
b
abc b
c
abc c
a
abc
abc
+
+
≤
+
+
+ +
+
+
tengsizlikni isbotlang.
90.
(Pol’sha -2000) Aytaylik
0 (
1,2,..., )
i
x
i
n
>
=
va
2
n
>
bo’lsin.
2
3
1
2
3
1
2
3
(
1)
2
3
...
...
2
n
n
n
n n
x
x
x
nx
x
x
x
x
−
+
+
+ +
≤
+ +
+
+ +
tengsizlikni isbotlang.
91.
(Gretsiya -2002) Agar , ,
a b c
musbat sonlar
2
2
2
1
a
b
c
+
+
=
shartni
qanoatlantirsa,
(
)
2
2
2
2
3
1
1
1 4
a
b
c
a a b b c c
b
c
a
+
+
≥
+
+
+
+
+
tengsizlikni
isbotlang.
92.
(Ukraina -2002)
1
2
1 (
1,2,..., ),
1,
1
...
i
n
a
i
n n
A
a
a
a
≥
=
≥
= + +
+ +
1
1
, 1
1
k
k k
x
k n
a x
−
=
≤ ≤
+
deb belgilash kiritsak, u holda
2
1
2
2
2
...
n
n A
x
x
x
n
A
+
+ +
>
+
tengsizlikni isbotlang.
93.
(Sankt Peterburg -2004) Musbat
, ,
a b c
sonlar uchun
2
20
27
3
2
2
49
ab
bc
ac
a
b
c
a b b
c c
a
+
+
+
+
≤
+
+
+
18
tengsizlikni isbotlang.
94.
(Irlandiya -1998)
0
x
≠
son uchun
8
5
4
1
1
0
x
x
x
x
−
− +
≥
tengsizlikni isbotlang.
95.
(Eron -1998) Birdan katta , ,
x y z
sonlar
1
1
1
2
x
y
z
+ + =
shartni qanoatlantirsa,
1
1
1
x y z
x
y
z
+ + ≥
− +
− +
−
tengsizlikni isbotlang.
96.
(Vyetnam-1998)
1
2
, ,...,
(
2)
n
x x
x n
≥
musbat sonlar
1
2
1
1
1
1
...
1998
1998
1998 1998
n
x
x
x
+
+ +
=
+
+
+
tenglikni qanoatlantirsa,
1 2
...
1998
1
n
n
x x x
n
≥
−
tengsizlikni isbotlang.
19
Yechimlar.
1.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan
munosabatga ko’ra,
( )
....
....
(
)
(
)
n
n k
nk
n k
n k
n
n
n
n
n k
k
k
nk
k
n
a
b
a
a
b
b
b
n k
n k a
b
b
b
+
+
+
+
+
+
+
+
+
+
≥
+
=
+
yoki
(
)
n k
h
n
k
a
n
k b
n k a
b
+
⋅
+ ⋅
≥
+
.
Xuddi shunday,
(
)
(
) ,
n k
n
n
k
n k
n
n
k
b
n
kc
n k b
c
c
n
k a
n k c
a
+
+
⋅
+
≥
+
⋅
+ ⋅
≥
+
tengsizliklarni hosil qilamiz. Bu tengsizliklarni hadma-had qo’shib,
n k
n k
n k
n
n
n
k
k
k
a
b
c
a
b
c
b
c
a
+
+
+
+
+
≥
+
+
ni hosil qilamiz.
2.
Ushbu
1 1
4
a b
a b
+ ≥
+
tengsizlikdan foydalanamiz:
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
1
4
4
4
2
2
2
.
2 2
2
2
1
1
1
a
b
c
b c
a b
b c
a c
a b
a c
b a c b a
c b
a c
a
b
c
⎛
⎞
⎛
⎞
⎛
⎞
+
+
=
+
+
+
+
+
≥
⎜
⎟
⎜
⎟
⎜
⎟
−
−
−
+
+
+
+
+
+
⎝
⎠
⎝
⎠
⎝
⎠
⎛
⎞
+
+
=
+
+
⎜
⎟
+ +
+ +
+
+
+
+
+
⎝
⎠
3.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan
foydalanib,
2
2
2
2
2
4
4
1
1
4
8 2
8
4
4
ab
ab
a
b
a
b
=
≤
≤
+
−
−
−
−
−
−
20
tengsizlikni hosil qilamiz. Xuddi shunday,
2
2
2
1
1
4
4
4
bc
b
c
≤
+
−
−
−
,
2
2
2
1
1
4
4
4
ac
a
c
≤
+
−
−
−
.
Bu tengsizliklarni hadma-had qo’shib,
2
2
2
1
1
1
1
1
1
4
4
4
4
4
4
ab
bc
ac
a
b
c
+
+
≤
+
+
−
−
−
−
−
−
tengsizlikni hosil qilamiz. Berilgan
4
4
4
3
a
b
c
+
+
=
shartdan
2
2
a
<
ekanligi kelib
chiqadi.Bundan quyidagi
(
)
4
2
2
2
2
1
5
1 (2
) 0
4
18
a
a
a
a
+
−
−
≥ ⇔
≤
−
tengsizlik o’rinli. Xuddi shunday,
4
4
2
2
1
5
1
5
,
4
18
4
18
b
c
b
c
+
+
≤
≤
−
−
tengsizliklar o’rinli. Bu tengsizliklarni hadma-had qo’shib,
4
4
4
2
2
2
1
1
1
1
1
1
5
5
5
1
4
4
4
4
4
4
18
18
18
a
b
c
ab
bc
ca
a
b
c
+
+
+
+
+
≤
+
+
≤
+
+
=
−
−
−
−
−
−
ekanligini hosil qilamiz.
4.
Berilgan tengsizlikni chap tomonida turgan qo’shiluvchilarni mos ravishda
A,B,C
deb belgilaymiz.
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
(
) (
)
(
)
(
)
(
)
2
1
2
2
y
z
by cz bz cy
b
c yz bc y
z
b
c
bc y
z
a
x
y
z
b c
A
b c
y
z
⎛
⎞
+
+
+
=
+
+
+
≤
+
+
+
=
⎜
⎟
⎝
⎠
⎛
⎞
=
+
+
⇒
≥ ⎜
⎟
+
+
⎝
⎠
Xuddi shunday ,
2
2
2
2
2
2
2
2
2
,
2
b
y
c
z
B
C
a c
t
x
a b
x
y
⎛
⎞
⎛
⎞
≥
≥
⎜
⎟
⎜
⎟
+
+
+
+
⎝
⎠
⎝
⎠
tengsizliklarni hosil
qilamiz.Berilgan shartlarga ko’ra
2
2
2
2
2
2
2
2
2
,
a
b
c
x
y
z
b c
c a
a b y
z
z
x
x
y
≥
≥
≥
≥
+
+
+
+
+
+
munosabatlar o’rinli. Chebishev tengsizligini qo’llasak,
21
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
2
3
1 1
2
3 3
a
b
c
x
y
z
A B C
b c
a c
a b
y
z
x
z
x
y
a
b
c
x
y
z
b c c a
a b
y
z
z
x
x
y
⎧
⎫⎧
⎫
⎪
⎪
⎛
⎞
⎛
⎞
⎛
⎞
+ + ≥ ⋅
+
+
+
+
≥
⎨
⎬⎨
⎬
⎜
⎟
⎜
⎟
⎜
⎟
+
+
+
+
+
+
⎝
⎠
⎝
⎠
⎝
⎠ ⎩
⎭
⎪
⎪
⎩
⎭
⎛
⎞
⎛
⎞
≥ ⋅ ⋅
+
+
+
+
⎜
⎟
⎜
⎟
+
+
+
+
+
+
⎝
⎠ ⎝
⎠
.
Musbat , ,
α β γ
sonlar uchun
3
2
α
β
γ
β γ γ α α β
+
+
≥
+
+
+
tengsizlikni isbotlaymiz.
,
,
s
t
α β τ
β γ
γ α
+ =
+ =
+ =
belgilash kiritib,
(
)
2
2
2
1
1
3
3
2 2 2 3
2
2
2
t s
s t
s t
s
t
t
s
s
t
s
s
t
t
α
β
γ
τ
τ
τ
β γ γ α α β
τ
τ
τ
τ τ
+ −
+ −
+ −
+
+
=
+
+
=
+
+
+
⎛
⎞
=
+ + + + + −
≥
+ + − =
⎜
⎟
⎝
⎠
ni hosil qilamiz. Bundan
2
1 1
3
3
3
2
3 3
2
2
4
A B C
⎛ ⎞
+ + ≥ ⋅ ⋅ ⋅
⋅ =
⎜ ⎟
⎝ ⎠
Tenglik
a=b=c
va
x=y=z
bo’lganda bajariladi.
5.
2
2
2
1
2
...
n
t a
a
a
=
+
+ +
deb belgilab,
(
)
(
)
{
}
(
)
1 2
2 3
1
1
1 2
1 3
1 4
1
2 3
2 4
2
3 4
3
1
2
2
2
2
1
2
1
2
...
...
...
...
.....
1
1
...
...
1
2
2
n n
n
n
n
n
n
n
n
n
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
a
a
a
a
a
a
t
−
−
+
+ +
+
≤
≤
+
+
+ +
+
+
+
+ +
+
+
+ +
+
+
+
+
=
=
+
+
−
+
+
=
−
munosabatlarni hosil qilamiz. Bu yerdan
t<
1
kelib chiqadi
.
Koshi-Bunyakovskiy-Shvarts tengsizligini qo’llab,
(
)
(
)(
)
2
2
2
2
2
2
2
1 1
2 2
1
2
1
2
2
...
...
...
n n
n
a b
a b
a b
a
a
a
b
b
b
t
+
+ +
≤
+
+ +
+
+ +
=
yoki
1 1
2 2
...
n n
a b
a b
a b
t
+
+ +
≤
tengsizlikni topamiz. Bundan
22
(
)
(
)
(
) (
)
(
)
(
)
(
)
1
1
2
2
2
3
1
1 1
2 2
2
1 2
2 3
1
...
...
1
1
....
1
1
1 1
2
2
n
n
n n
n
a b
a
a b
a
a b
a
a b
a b
a b
a a
a a
a a
t
t
t
+
+
+
+ +
+
=
+
+ +
+
+
+
+
+
≤
+
− = −
−
+ <
|