Induktsiya bazasi
.
1
n
=
da:
1
1
1
1
13
...
1
1 1 1 2
3 1 1 12
S
=
+
+ +
=
>
+
+
⋅ +
.
Induktsiya bazasi isbotlandi.
Induktiv o’tish.
n k
=
da
1
1
1
...
1
1
2
3
1
k
S
k
k
k
=
+
+ +
>
+
+
+
tengsizlik
bajariladi deb faraz qilamiz.
1
1
1
1
1
1
1
...
1
2
3
3
1 3
2 3
3 3
4
k
S
k
k
k
k
k
k
+
=
+
+ +
+
+
+
+
+
+
+
+
+
>
tengsizlik bajarilishini isbotlash kerak.
1
1
1
1
1
1
1
1
1
...
2
3
3
1 3
2 3
3 3
4
1
1
k
S
k
k
k
k
k
k
k
k
+
⎛
⎞
=
+
+ +
+
+
+
+
−
⎜
⎟
+
+
+
+
+
+
+
+
⎝
⎠
=
25
1
1
0
1
1
1
1
1
1
1
1
...
1
1
2
3
3
1 3
2 3
3 3
4
1
S
k
k
k
k
k
k
k
k
= >
>
=
+
+
+ +
+
+
+
−
>
+
+
+
+
+
+
+
+
.
1
1
1
1
1
1
2
3
2 3
3 3
4
1 3
2 3
4 3
3
k
k
k
k
k
k
k
+
+
−
=
+
−
+
+
+
+
+
+
+
=
(3
4)(3
3) (3
2)(3
3) (6
4)(3
4)
2
0
(3
2)(3
3)(3
4)
(3
2)(3
3)(3
4)
k
k
k
k
k
k
k
k
k
k
k
k
+
+ +
+
+ −
+
+
=
=
+
+
+
+
+
+
>
"
ekanligidan
tengsizlik kelib chiqadi. Matematik induktsiya printsipiga
asoslanib, ixtiyoriy natural son uchun (10) tengsizlik bajariladi deb xulosa qilamiz.
" 0
>
n
6-masala.
2
2
2
x
y
z
xy xz yz
+
+
≥
+
+
tengsizlikni isbotlang, bu yyerda
, ,
x y z
-
musbat sonlar.
Yechilishi.
Ma’lum
2
2
2
2
2
2
2 ,
2 ,
2
x
y
xy x
z
xz y
z
+
≥
+
≥
+
≥
yz
)
tengsizliklarni
qo’shib, ushbu
2
2
2
2
2
2
2
2
2
(
) (
) (
) 2(
) 2(
x
y
x
z
y
z
x
y
z
xy xz yz
+
+
+
+
+
≥
+
+
≥
+
+
tengsizlikni olamiz.
7-masala.
4
4
4
(
)
x
y
z
xyz x y z
+
+
≥
+ +
tengsizlikni isbotlang, bu yerda , ,
x y z
-
musbat sonlar.
Yechilishi.
1-masalaga ko’ra:
4
4
4
2 2
2 2
2 2
2 2
2 2
2
( )
( )
( )
2
x
y
z
x
y
z
x y
y z
x z
+
+
=
+
+
≥
+
+
ga egamiz. Bu yerdan esa
2
2
2 2
2 2
(
)
x y
y z
x z
xyyz yzzx zxxy xyz x y z
+
+
≥
+
+
=
+ +
ni olamiz.
8-masala.
4
4
4
4
4
x
y
z
u
xyzu
+
+
+
≥
tengsizlikni isbotlang, bu yerda , , ,
x y z u
-
musbat sonlar.
Yechilishi.
4
4
2 2
4
4
2
2
,
2
2
x
y
x y
z
u
z
+
≥
+
≥
u
2
ga egamiz. Demak,
4
4
4
4
2 2
2
2
2
x
y
z
u
x y
z u
+
+
+
≥
+
. Bundan tashqari
2 2
2 2
2
x y
z u
xyz
+
≥
u
. Demak,
4
4
4
4
4
x
y
z
u
xyzu
+
+
+
≥
.
26
9-masala.
2
1
1
(
)
(
)
2
4
x y
x y
x y
y
+
+
+
≥
+
x
tengsizlikni isbotlang, bu yerda
,
x y
- musbat sonlar.
Yechilishi.
Birinchidan,
2
1
1
1
(
)
(
)
(
)(
2
4
2
x y
x y
x y x y
1
)
2
+
+
+
=
+
+ +
.
Ikkinchidan,
2
x y
xy
+
≥
,
1
1
1
2
4
4
x y
x
y
x
+ + = + + + ≥
+
y
.
Demak,
2
1
1
(
)
(
)
(
)
2
4
x y
x y
xy
y
x
x y
y
+
+
+
=
+
≥
+
x
.
10-masala.
0,
0
x
y
≥
≥
va
2
x y
+ =
bo’lsin.
2 2
2
2
(
)
x y x
y
+
≤
2
1
tengsizlikni isbotlang.
Aniqlik uchun
1
,
1
, 0
x
y
ε
ε
ε
= +
= −
≤ ≤
deb olamiz. U holda
2 2
2
2
2
2
2
2
2 2
2
(
) (1
) (1
) ((1
)
(1
) ) (1
) (2 2 )
x y x
y
ε
ε
ε
ε
ε
ε
+
= −
+
−
+ +
= −
+
=
2
2
2
2
4
2(1
)(1
)(1
) 2(1
)(1
) 2
ε
ε
ε
ε
ε
=
−
−
+
=
−
−
≤
11-masala
. va
b
bir xil ishorali sonlar bo’lsin.
a
2 2
2
2
2
3
(
)
10
4
12
a b a b
a
ab b
+
+
≤
+
ekanligini isbotlang.
Qachon tenglik bajariladi?
ekanligini hisobga olib va Koshi
tengsizligidan foydalanib quyidagiga ega bo’lamiz:
0
ab
>
2
2
2
2
3
2
(
)
10
4
2
3
1
a
ab b
ab ab
a b
a
ab b
ab ab
+
+
+
+
+
+
⋅
⋅
≤
=
2
+
.
Tenglik esa
bo’lganda bajarilishini eslatib o’tamiz.
a b
=
12-masala
. va birdan katta sonlar bo’lsin.
,
a b
c
2
2
2
log (
)log (
)log (
) 1
a
b
b
b
c
a
b ac
c ab
a bc
ac
ab
bc
− +
− +
− +
≥
tengsizlikni isbotlang.
27
1,
1,
1
a
b
c
>
>
>
va
2
2
2
2 ,
2 ,
2
b
c
a
ac
b
ab
c
bc
a
ac
ab
bc
+
≥
+
≥
+
≥
bo’lgani uchun
2
2
2
log (
)log (
)log (
) log (2
)
a
b
b
a
b
c
a
b ac
c ab
a bc
b b
ac
ab
bc
− +
− +
− +
≥
− ×
=
.
log (2
)log (2
) log
log log
1
b
c
a
b
c
c c
a a
b
c
a
×
−
−
=
13-masala
. va
b
musbat sonlar bo’lsin.
a
3
3
3
1 1
2(
)(
)
a
b
a b
b
a
a b
+
≤
+
+
tengsizlikni isbotlang.
Yechilishi.
Berilgan tengsizlikni kubga ko’tarish va soddalashtirishlardan so’ng
quyidagiga ega bo’lamiz:
3
3
3(
) 4
a
b
a
b
a
b
+
≤ + +
b
a
. Koshi tengsizligiga ko’ra
3
1 1
3
a
b
b
+ + ≥
a
va
3
1 1
3
b
a
a
+ + ≥
b
tengsizliklarni olamiz va ularni qo’shib, qidirilayotgan tengsizlikni
olamiz.
14-masala.
2
1
2
1
3
1
(
3
1
4(
1)
n
k n
n
n
n N
n
k
n
= +
)
+
≤
≤
∈
+
+
∑
ekanligini isbotlang.
Yechilishi.
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
3
1
(
)
3
1
2
3
1
2
(3
1
)
n
n
n
n
k n
k n
k n
k n
n
k
n
k
k
n
k
k
k n
= +
= +
= +
= +
+
=
=
+
=
≤
+ −
+ −
+ −
∑
∑
∑
∑
k
;
1)
2
2
2
(3
1)
3
1
(3
1)
(3
1
)
(
)
4
4
4
n
n
n
k n
k
k
+
+
+
+ −
=
−
−
≤
tengsizlikdan
2
2
1
1
3
1
4 (3
1)
2
2
(3
1
) 2(3
1)
3
n
k n
n
n n
k n
k
n
n
= +
+
+
≥
=
+ −
+
+
∑
1
n
ekanligi kelib chiqadi;
28
2)
(3
1
) 2 (
1) (2
)(
(
1)) 0 (
1
2 )
k n
k
n n
n k k
n
n
k
n
+ −
−
+ =
−
− +
≥
+ ≤ ≤
tengsizlikdan
2
1
1
3
1
(3
1)
3
2
(3
1
) 4 (
1) 4(
1
n
k n
n
n n
n
k n
k
n n
n
= +
1
)
+
+
+
≤
=
+ −
+
+
∑
ekanligi kelib chiqadi.
Demak
2
1
2
1
3
1
(
)
3
1
4(
1)
n
k n
n
n
n N
n
k
n
= +
+
≤
≤
∈
+
+
∑
.
15-masala
.
va musbat sonlar bo’lsin.
tengsizlik
va faqat biror uchburchak tashkil qilgandagina bajarilishi
mumkinligini isbotlang.
,
a b
c
4
4
4
2
2
2
2(
) (
)
a
b
c
a
b
c
+
+
<
+
+
2
0
=
,
a b
c
Yechilishi.
Ravshanki, bizning tengsizligimiz
4
4
4
2 2
2 2
2 2
2
2
2
a
b
c
a b
b c
a c
+
+
−
+
+
<
tengsizlikka teng kuchli. Oxirgi
tengsizlikning chap tomonini almashtiramiz:
4
4
4
2 2
2 2
2 2
2
2
2
2 2
2
2
2
2
2
2
2
2
2
2
2
2
2
4
(
2 )(
2 ) ((
)
)((
)
)
(
)(
)(
)(
)
a
b
c
a b
b c
a c
a
b
c
a b
a
b
c
ab a
b
c
ab
a b
c
a b
c
a b c a b c a b c a b c
+
+
−
+
+
=
+
− −
=
=
+
− −
+
− +
=
−
−
+
−
=
− +
− −
+ +
+ −
Demak, berilgan tengsizlik
va biror uchburchak tashkil qilganda aniq
ravishda bajariladigan ushbu
,
a b
c
(
)(
)(
)(
) 0
a b c a b c a b c a b c
− +
− −
+ +
+ − >
tengsizlikka teng kuchli.
Endi faraz qilamiz, bu tengsizlik bajariladi, biroq
va biror uchburchak
tashkil qilmaydi. U holda
,
a b
c
,
,
,
a b c a b c a b c a b c
− +
− −
+ +
+ −
sonlardan kamida
ikkitasi manfiy.
va
0
a b c
+ − <
0
b c a
+ − <
bo’lsin. Bu yerdan masalaning shartiga
zid bo’lgan
tengsizlikni olamiz.
2
0
b
<
16-masala
.
–
ketma-ketlikning biror o’rin
almashtirishi bo’lsin.
1
2
, , ... ,
n
b b
b
1
2
,
, ... ,
n
a a
a
1
2
1
2
1
1
1
(
)(
)...(
) 2
n
n
n
a
a
a
b
b
b
+
+
+
≥
29
tengsizlikni isbotlang.
Tengsizlikning isboti
Koshi tengsizligidan darhol kelib chiqadi:
1 2
1
2
1
2
1
2
1
2
1 2
...
1
1
1
(
)(
)...(
) 2
2
...2
2
...
n
n
n
n
n
n
a
a a a
a
a
a
a
a
b
b
b
b
b
b
b b
b
+
+
+
≥
=
=
n
.
17-masala.
10
6
5
3
2
1 0
x
x
x
x
x
x
+
+
+
+
+ + >
|
