-
misolda 
lemma - Sonli tengsizliklar haqida. Toshkent 2008

69-
misolda
lemma
sifatida isbotlangan.
80.
Avvaliga quyidagi lemmani isbotlaymiz:
Lemma. Musbat ,
x y
sonlar uchun
7
7
3
3
5
5
x
y
x
y
x
y
x y
+
+
≥
+
+
munosabat o’rinli.
Lemmaning isboti: Haqiqatdan ham
(
)
(
)
(
)(
) (
) (
)
(
)
(
) (
)(
)
(
) (
)
7
7
3
3
5
5
7
5 3
7
5 3
2
5
2
2
5
2
2
2
2
5
5
2
2
2
2
0
x
y
x y
x
y
x
y
x y x y
y x y x
x y x
y
y x x
y
x
y
x y y x
xy x
y
x
y
+
+
−
+
+
=
−
+
−
=
=
−
−
−
=
−
−
=
−
+
≥
tenglik
x y
=
bo’lganda bajariladi.
Lemmadan foydalansak,
(
) (
) (
) (
)
(
)
7
7
7
7
7
7
3
3
3
3
3
3
5
5
5
5
5
5
2
2
2
2
2
2
2
2
2
2
a
b
b
c
c
a
a
b
b
c
c
a
a
b
b
c
c
a
a b
b c
c a
a
ab b
b
bc c
c
ac a
a
b
c
ab bc ca
+
+
+
+
+
+
+
+
≥
+
+
=
+
+
+
+
+
+
=
−
+
+
−
+
+
−
+
=
+
+
−
+
+
munosabat hosil bo’ladi.
Endi
(
)
(
)
2
2
2
1
2
3
a
b
c
ab bc ca
+
+
−
+
+
≥
ekanligini ko’rsatamiz. Haqiqatdan
ham
(
)
(
)
(
)
2
2
2
2
2
2
2
1
2
3
3
a b c
a
b
c
ab bc ca
a
b
c
+ +
+
+
−
+
+
≥
+
+
≥
=
.
52
81.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligini
quyidagi usulda qo’llaymiz:
7
2
1
1
1
1
2
4
4
2
2
2
1
1
1
7
7
4
4
2 2 2
2
x y
x y
x
y
x y
x y
xy
x y
x y
xy
x y x y
x y
x y x y
xy
+
+
+ +
+
=
+
+
+
+ + +
≥
+
+
+
+
+
≥
⋅
⋅
⋅
⋅ ⋅ ⋅
=
+
+
82.
Avvaliga quyidagi
2
1
)
(
x
x
x
f
+
=
,
[ ]
0;1
x
∈
funktsiyani hossalarini o’rganamiz.
Ko’rinib turibdiki, ushbu funktsiya ko’rsatilgan oraliqda qavariq funktsiyadir. U
holda qavariq funktsiyalar uchun ushbu ( )
( )
( ) 3
3
x y z
g x
g y
g z
g
+ +
⎛
⎞
+
+
≤
⎜
⎟
⎝
⎠
Iensen
tensizligidan foydalanib,
2
2
2
1
9
( )
( )
( ) 3
3
1
1
1
3
3
10
a
b
c
a b c
f a
f b
f c
f
f
a
b
c
+ +
⎛
⎞
⎛ ⎞
+
+
=
+
+
≤
=
=
⎜
⎟
⎜ ⎟
+
+
+
⎝
⎠
⎝ ⎠
munosabatni hosil qilamiz.
83.
Umumiylikni chegaralamasdan
a b c d
≥ ≥ ≥
va
2
2
2
2
a
b
c
d
≥
≥
≥
deymiz. U
holda Chebishev tengsizligini qo’llab, quyidagi
(
)
(
)
(
) (
)
2
2
2
2
2
2
2
2
3
3
3
3
4
a
b
c
d
a b c d a
b
c
d
a
b
c
d
+
+
+
=
+ + +
+
+
+
≤
+
+ +
yoki
(
) (
)
3
3
3
3
2
2
2
2
3
6
2
a
b
c
d
a
b
c
d
+
+ +
≥
+
+
+
(*) munosabatni hosil qilamiz.
Endi Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda qo’llasak,
(
)
(
) (
)
2
2
2
2
2
1 1 1 1
a
b
c
d
a b c d
+
+
+
+ + + ≥
+ + +
yoki
(
)
2
2
2
2
1
1
2
8
a
b
c
d
+
+
+
≥
(**) munosabat hosil bo’ladi. (*) va (**) larni hadma-had qo’shish natijasida
yuqoridagi isboti talab etilgan tengsizlik kelib chiqadi.
53
84.
Istalgan natural
n
uchun
1
1
2
1 2
n
n
>
−
ekanligini etiborga olsak,
1 1 1
1
1 1
1
...
...
2 3 5
2
1 2 4
2
n
n
+ + + +
> + + +
−
munosabat o’rinlidir. Endi
1 1
1
1 1
1
...
...
2 2
2
2 4
2
n
n
+ + + > + + +
yoki
1
1 1
1 1
...
2
2 4
2
n n
⎛
⎞
>
+ + +
⎜
⎟
⎝
⎠
ekanligidan
foydalansak, U holda
1
1
1 1 1
1
1 1
1 1
1 1
1
1
...
...
...
...
3
2
1 2 2 3
2
1
2 4
2
2 4
2
1 1 1
1
...
2 4
2
n
n
n n
n
n
n
n
⎛
⎞
⎛
⎞
+ + +
= + + + +
>
+ + +
+
+ + +
=
⎜
⎟
⎜
⎟
−
−
⎝
⎠
⎝
⎠
+ ⎛
⎞
=
+ + +
⎜
⎟
⎝
⎠
bo’ladi.
85.
Berilgan tengsizlikda quyidagicha shakl almashtirish bajaramiz.
2
2
2
2
2
2
1
(1
)
1
1
a
b
a
b
ab
a
b
+
− −
+
≤
+
−
−
yoki
2
2
1
1
ab
a
b
≤
+
−
−
. Bu yerda
,
sin
,
0;
2
a tg
b
π
α
β α β
⎛
⎞
=
=
∈⎜
⎟
⎝
⎠
belgilash olsak, u holda
1
sin
cos
cos
tg
α
β
β
α
≤
−
yoki
(
)
cos
1
α β
−
≤
bo’ladi.
86.
Yuqoridagi berilgan shartlarga ko’ra quyidagi tengliklarni yozamiz:
2
1
1
3
2
2
1
1
2 2
2 2
3
2 3
2 3
3
.............................
2
2
3
n
n
n
x
x
x
x
x
x
n x
n x
x
−
−
⋅ ⋅ = ⋅ ⋅ −
⋅ ⋅ = ⋅ ⋅ −
⋅ ⋅
= ⋅ ⋅
−
va bu tengsizliklarni hadma-had qo’shib,
1
1
1
1
1
2
1
3
2 2
2
2
1 2
2
n
n
n
i
i
n
i
n
i
i
i
x
x
x
n x
x
n x
−
−
−
−
=
=
= ⋅ ⋅ +
− ⋅ ⋅
= +
− ⋅ ⋅
∑
∑
∑
yoki
1
1
0
n
n
i
i
nx
x
=
= −
>
∑
munosabatni hosil qilamiz.
54
87.
O’rta arifmetik va o’rta geometrik miqdorlar haqidagi Koshi tengsizligi va
0;
2
x
π
⎛
⎞
∈⎜
⎟
⎝
⎠
uchun
(
)
sin
sin 2
1
x
x
a
− ≤
yoki sin
2
tgx
x
≤
tengsizliklarni o’rinli
ekanligini etiborga olib,
1
2
1
2
1
2
2
1
2
2
2
sin
sin
... sin
sin
sin
... sin
...
2
...
2
2
n
n
n
n
n
n
n
n
n
n
x
x
x
x
x
x
n
tgx
tgx
tgx
n
tgx
tgx
tgx
n
−
−
−
+
+ +
⎛
⎞
⋅
⋅ ⋅
≤
≤
⎜
⎟
⎝
⎠
⎛
⎞
+
+ +
≤
⋅
≤
⎜
⎟
⎜
⎟
⎝
⎠
⎛
⎞
+
+ +
≤
⋅
≤
⎜
⎟
⎝
⎠
munosabatni hosil qilamiz.
88.
Umumlashgan Koshi-Bunyakovskiy-Shvarts tengsizligini quyidagi usulda
qo’llab,
(
)(
)
(
)
6
6
6
3
2
2
2
1 1 1
a
b
c
x y z
a
b
c
x
y
z
⎛
⎞
+
+
+ +
+ + ≥
+
+
⎜
⎟
⎝
⎠
munosabatni hosil
qilamiz. Bundan yuqoridagi isboti talab etilgan tangsizlik kelib chiqadi.
89.
Ushbu
(
)
3
3
x
y
xy x y
+
≥
+
tengsizlikdan foydalanamiz:
3
3
3
3
3
3
1
1
1
1
1
(
)
(
)
1
1
1
1
1
1
(
)
a
b
abc b
c
abc c
a
abc
ab a b
abc bc b c
abc
ca c a
abc
a b c ab bc ca
abc
+
+
≤
+
+
+
+
+ +
+
+
+ +
+ +
⎛
⎞
+
=
+
+
=
⎜
⎟
+
+
+ + ⎝
⎠
90.
I-usul:
Ushbu
(
1)
1 2 ... (
1)
2
n n
n
−
+ + +
− =
tenglikdan va o’rta arifmetik va
o’rta geometrik miqdorlar haqidagi Koshi tengsizligidan foydalanamiz:
55
(
)
2
3
2
3
1
2
3
1
2
3
2
3
1
2
3
1
1
2
3
(
1)
...
(1
) (2
) ...
(
1)
2
( ) (1
) (1 1
) ... (1 1 ... 1
)
2
3
...
n
n
n
n
n
n
n
n
n n
x
x
x
x
x
x
x
n
x
x
x
x
x
x
x
x
nx
−
−
+ +
+
+ +
= + +
+
+
+ +
− +
=
=
+ +
+ + +
+ + + + + +
≥
≥ +
+
+ +
(
)
(
)
(
)
2
2
1
2
1
2
1
2
...
(
1) 1
...
(
1) 1
1 2
1
n
n
n
n
x
x
x
x
x
x
x
x
+
+ +
= +
− +
+ +
− +
≥ + +
−
II-usul:
Bernulli tengsizligidan quyidagi usulda foydalansak:
(
) (
)
(
)
(
)
(
)
(
)
3
2
2
3
1
2
3
1
2
3
1
2
3
1
2
3
(
1)
...
(
1) 1
(
1) 1
...
(
1) 1
2
(
1)
(
1)
1 2(
1)
1 3(
1)
...
1
(
1)
2
2
2
3
...
n
n
n
n
n
n
n n
x
x
x
x
x
x
x
x
n n
n n
x
x
x
n x
x
x
x
nx
−
+
+
+ +
+
= +
− +
+
− +
+ +
− +
+
−
−
+
≥ + +
−
+ +
−
+ + +
−
+
=
= +
+
+ +
munosabat hosil bo’ladi.
91.
Bu tengsizlikni chap qismini
S
bilan belgilab, quyidagi usulda Koshi-
Bunyakovskiy-Shvarts tengsizligini qo’llaymiz:
(
) (
) (
)
(
)
(
)
2
2
2
2
2
2
2
1
1
1
S a b
b c
c a
a a b b c c
+ +
+ +
+
≥
+
+
bundan,
(
)
2
2 2
2 2
2 2
1
a a b b c c
S
a b
b c
c a
+
+
≥
+
+
+
tengsizlikni va undan
(
) (
)
(
)
(
)
2
2
2
2
2 2
2 2
2 2
2
2
2
3
1
4
1
3
a a b b c c
a a b b c c
S
a a b b c c
a b
b c
c a
a
b
c
+
+
+
+
≥
=
+
+
+
+
+
+
+
+
munosabatni hosil qilamiz.
92.
1
k
k
y
x
=
almashtirish olsak, u holda,
56
1
1
1
1
1
.
1
k
k
k
k
k
k
k
a
x
y
a
y
y
y
−
−
=
=
⇔
= +
+
1
1,
1
k
k
y
a
−
≥
≥
ekanligidan
(
)
1
1
1
1
1
1
1
1
0
1
k
k
k
k
k
k
a
a
a
y
y
y
−
−
−
⎛
⎞
−
− ≤ ⇔ +
≤
+
⎜
⎟
⎝
⎠
bundan
1
1
1
1
k
k
k
k
k
a
y
a
y
y
−
−
= +
≤
+
munosabatni hosil qilamiz.
1
1
1
1
1
1
1
1
1
1
0
1
1
1
1
1
n
n
n
n
n
n
n
k
k
k
k
k
k
k
k
k
k
k
k
k
k
y
a
a
A
A
y
y
y
y
y
−
−
=
=
=
=
=
=
=
−
≤
+
=
+
+
= +
< +
∑
∑
∑
∑
∑
∑
∑
1
1
n
k
k
t
y
=
=
∑
deb belgilash kiritamiz, bundan
2
1
,
0
n
k
k
n t
y
t
=
≥
>
∑
tengsizlikni hosil qilamiz.
2
2
2
2
2
2
2
2
1
2
2
2
2
2
2
2
4
2
0
2
4
2
2
4
4
2
n
k
k
n
A
A
n
n
y
A t
t
At n
t
t
A
A
n
n
n
n A
n
A
n
n
A A
A A
A
A
A
A
=
− +
+
≤
< + ⇔ +
−
≥ ⇔ >
=
=
+
+
=
≥
+
+
⎛
⎞
+ +
+
+
⎜
⎟
+
⎝
⎠
∑
93.
12
3
49
ab
b a
a b
+
≤
+
chunki
2
8
9
2(
2 )
0,
2
49
bc
b
c
a
b
b
c
+
−
≥
≤
+
chunki
(
)
2
18
2 2
3
0,
2
49
ac
c a
b
c
c
a
+
−
≥
≤
+
chunki
(
)
2
2 3
0
c a
−
≥
, bu tengsizliklarni hadma-
had qo’shib, isbotlash kerak bo’lgan tengsizlikni hosil qilamiz.
tenglik 2
3
a
b
c
=
=
bo’lganda bajariladi.
94.
9
3
1,
1
x
x
−
−
ifodalar ishorasi bir xil hamda
4
0
x
>
bo’lgani uchun
3
9
3
8
5
5
3
4
4
4
1
1
1 (
1)(
1)
(
1)
0
x
x
x
x
x
x x
x
x
x
x
−
−
−
−
− +
=
− −
=
≥
.
57
95.
Ravshanki,
1
1
1
1
x
y
z
x
y
z
−
−
−
+
+
=
.
Koshi-Bunyakovskiy-Shvarts tengsizligiga ko’ra
1
1
1
1
1
1
x
y
z
x y z
x
y
z
x
y
z
−
−
−
+ +
+
+
≥
− +
− +
−
.
96.
1998
1998
i
i
y
x
=
+
almashtirish kiritamiz.
Ravshanki, 0,
1,2,...,
i
y
i
n
≥
=
va
1
2
...
1
n
y
y
y
+
+ +
=
.
Demak, 1
i
j
j i
y
y
≠
− =
∑
.
Koshi tengsizligiga ko’ra
1
1
(
1)
n
i
j
j i
y
n
y
−
≠
− ≥
−
∏
.
Bu tengsizliklarni barchasini ko’paytirsak,
1
1
(1
) (
1)
n
n
n
i
i
i
i
y
n
y
=
=
−
≥
−
∏
∏
yoki
1
1
(
1)
n
n
i
i
i
y
n
y
=
−
≥
−
∏
tengsizlikni hosil qilamiz.
1
1998
i
i
i
y
x
y
−
=
bo’lgani uchun bundan
1 2
...
1998 (
1)
n
n
n
x x x
n
≥
−
tengsizlikni hosil
qilamiz.
58
Manbaalar ro’yxati
1.
Hojoo Lee. Topics in Inequalities-Theorems and Techniques. Seoul: 2004.
2.
Andreescu T., Dospinescu G., Cirtoaje V., Lascu M. Old and new inequalities.
Gil Publishing House, 2004.
3.
Mathematical Olympiads, Problems and solutions from around the world, 1998-
1999. Edited by Andreescu T. and Feng Z. Washington. 2000.
4.
Math Links, http://www.mathlinks.ro
5.
Art of Problem Solving, http://www.artofproblemsolving.com
6.
Math Pro Press, http://www.mathpropress.com
7.
K.S.Kedlaya, Aindex.html
8.
T.J.Mildorf, Olympiad Inequalities,
http://web.mit.edu/tmildorf
9.
Математические
задачи
,
http://www.problems.ru
10.
«
Математика
в
школе
» (
Россия
), «
Квант
» (
Россия
), «
Соровский
образовательный
журнал
» (
Россия
), “Crux mathematicorum with
mathematical Mayhem” (
Канада
), “Fizika, matematika va informatika”
(
Ўзбекистон
)
журналлари
.
59
MUNDARIJA
Masalalar…………...............................................................................
3
Yechimlar……………..........................................................................
19
Manbaalar
ro’yhati………….................................................................
58
Document Outline - §1. Sonli tengsizliklar haqida.
-
- §2. O’rtacha qiymatlar va ular orasidagi munosabatlar.
- 1-BOB. FUNKSIYANING XOSSALARI YORDAMIDA TENGSIZLIKLARNI ISBOTLASH USULLARI
- 1-§. Funksiyaning monotonlik xossasi yordamida isbotlanadigan tengsizliklar
-
- 2-§. Funksiyaning qavariqlik xossasi yordamida isbotlanadigan tengsizliklar
- Isbot.
-
- funksiya uchun Yensen tengsizligini yozamiz.
- , .
|