Sonli tengsizliklar va ularning xossalari. tengsizliklarni isbotlash

Информация о документе

Цена	12000UZS
Размер	1.2MB
Покупки	0
Дата загрузки	07 Ноябрь 2026
Расширение	pdf
Раздел	Дипломные работы
Предмет	Педагогика

Sonli tengsizliklar va ularning xossalari. tengsizliklarni isbotlash

Купить

2. Sonli tengsizlik larning asosiy xossalari

1.1 -m iso l. Agar &#3627408475;
&#3627408474; to’g’ri kasr bo’lsa, u holda &#3627408475;
&#3627408474; < &#3627408475;+1
&#3627408474;+1 bo’lishini isbotlang.
∆ &#3627408475;
&#3627408474; kasr &#3627408475; < &#3627408474; bo’lganda (n va m – natural sonlar) to’g’ri kasr deb ata lishini
eslatib o’tamiz. Ushbu &#3627408475;
&#3627408474; − &#3627408475;+1
&#3627408474;+1= &#3627408475;(&#3627408474;+1)−&#3627408474;(&#3627408475;+1)
&#3627408474;(&#3627408474;+1) = &#3627408475;−&#3627408474;
&#3627408474;(&#3627408474;+1) ayirma noldan kichik ,
chunki &#3627408475; − &#3627408474; < 0,&#3627408474; > 0,&#3627408474; + 1 > 0. Binobarin, &#3627408475;
&#3627408474; < &#3627408475;+1
&#3627408474;+1 .▲

1. 2-m iso l. Agar &#3627408462; ≠ &#3627408463; bo’lsa, u holda &#3627408462;2+ &#3627408463;2 > 2&#3627408462;&#3627408463; bo’lishini isbotlang.
∆ &#3627408462;2+ &#3627408463;2− 2&#3627408462;&#3627408463; ayirma musbat ekanligini isbotlaymiz. Haqiqatan ham, &#3627408462;2+
&#3627408463;2− 2&#3627408462;&#3627408463; = (&#3627408462; − &#3627408463;)2 > 0, chunki &#3627408462; ≠ &#3627408463; . ▲

1. 1-teorema . Agar &#3627408462; > &#3627408463; va &#3627408463; > &#3627408464; bo’lsa, u holda &#3627408462; > &#3627408464; bo’ladi.
Δ Shartga ko’ra &#3627408462; > &#3627408463; va &#3627408463; > &#3627408464;. Bu &#3627408462; − &#3627408463; > 0 va &#3627408463; − &#3627408464; > 0 ekanini bildiradi. &#3627408462; −
&#3627408463; va &#3627408463; − &#3627408464; musbat sonlarni qo’shib, (&#3627408462; − &#3627408463;)+ (&#3627408463; − &#3627408464;)> 0 ni hosil qilamiz, ya’ni
&#3627408462; − &#3627408464; > 0. Demak, &#3627408462; > &#3627408464;. ▲

1.2 -teorema. Agar tengsizlikning ikkala qismiga ayni bir son qo’shilsa, u holda
tengsizlik ishorasi o’zgarmaydi.
Δ &#3627408514; > &#3627408515; bo’lsin . Bu holda ixtiyoriy &#3627408464; son uchun &#3627408462; + &#3627408464; > &#3627408463; + &#3627408464; teng -sizlikning
bajarilishini isbotlash talab qilinadi. Ushbu (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408464;)= &#3627408462; + &#3627408464;− &#3627408463; − &#3627408464; =
&#3627408462; − &#3627408463;ayirmani qaraymiz. Bu ayirma musbat, chunki masalaning shartiga ko’ra &#3627408462; >
&#3627408463;. Demak, &#3627408462; + &#3627408464; > &#3627408463; + &#3627408464;. ▲

1.1 -Natija. Istalgan qo’shiluvchini tengsizlikning bir qismidan ikkinchi qis -miga
shu qo’shiluvchining ishorasini qarama -qarshisiga almashtirgan holda ko’chirish
mumkin.
Δ &#3627408514; > &#3627408515; + &#3627408516; bo’lsin . Bu tengsizlikning ikkala qismiga – &#3627408464; sonni qo’shib, &#3627408462; − &#3627408464; >
&#3627408463; + &#3627408464; − &#3627408464; ni hosil qilamiz, ya’ni &#3627408462; − &#3627408464; > &#3627408463; ▲

1. 3-teorema. Agar tengsizlikning ikkala qismi ayni bir musbat songa ko’ paytrilsa,
u holda tengsizlik ishorasi o’zgarmaydi. Agar tengsizlikning ikkala qismi ayni bir
manfiy songa ko’paytrilsa,u holda tengsizlik ishorasi qarama -qarshisiga o’zgaradi.
Δ 1) &#3627408514; > &#3627408515; va &#3627408516; > ?????? bo’lsin . &#3627408462;&#3627408464; > &#3627408463;&#3627408464; ekanini isbotlaymiz.
Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464; > 0. Shuning uchun (&#3627408462; − &#3627408463;)&#3627408464; > 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408464; >
0. Demak, &#3627408462;&#3627408464; > &#3627408463;&#3627408464; .
2) &#3627408514; > &#3627408515; va &#3627408516; < ?????? bo’lsin . &#3627408462;&#3627408464; < &#3627408463;&#3627408464; ekanini isbotlaymiz. Shartga ko’ra &#3627408462; − &#3627408463; > 0
va &#3627408464; < 0. Shuning uchun (&#3627408462; − &#3627408463;)&#3627408464; < 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408464; < 0. Demak, &#3627408462;&#3627408464; < &#3627408463;&#3627408464; . ▲
1. 2-Natija. Agar tengsizlikning ikkala qismi ayni bir musbat songa bo’linsa, u holda
tengsizlik ishorasi o’zgarmaydi. Agar tengsizlikning ikkala qismi ayni bir manfiy
songa bo’linsa,u holda tengsizlik ishorasi qarama -qarshisiga o’zga -radi.

1.4. -teorema. Bir xil ishorali tengsizliklarni qo’shishda xuddi shu ishorali tengsizlik
hosil bo’ladi: agar &#3627408462; > &#3627408463; va &#3627408464; > &#3627408465; bo’lsa, u holda &#3627408462; + &#3627408464; > &#3627408463; + &#3627408465; bo’ladi.
Δ Shartga ko’ra &#3627408462; − &#3627408463; > 0 va &#3627408464;− &#3627408465; > 0. Ushbu ayirmani qaraymiz: (&#3627408462; + &#3627408464;)−
(&#3627408463; + &#3627408465;)= &#3627408462; + &#3627408464; − &#3627408463; − &#3627408465; = (&#3627408462; − &#3627408463;)+ (&#3627408464;− &#3627408465;). Musbat sonlarning yig’indisi
musbat bo’lgani uchun (&#3627408462; + &#3627408464;)− (&#3627408463; + &#3627408465;)> 0,ya’ni (&#3627408462; + &#3627408464;)> (&#3627408463; + &#3627408465;). ▲
1. 5. -teorema. Chap va o’ng qismlari musbat bo’lgan bir xil ishorali teng -sizliklarni
ko’paytirish natijasida xuddi shu ishorali tengsizlik hosil bo’ladi: agar &#3627408462; > &#3627408463;,&#3627408464; >
&#3627408465; va &#3627408462;,&#3627408463;,&#3627408464;,&#3627408465;– musbat sonlar bo’lsa , u holda &#3627408462;&#3627408464; > &#3627408463;&#3627408465; bo’ladi.
Ushbu ayirmani qaraymiz:
&#3627408462;&#3627408464; − &#3627408463;&#3627408465; = &#3627408462;&#3627408464; − &#3627408463;&#3627408464; + &#3627408463;&#3627408464; − &#3627408463;&#3627408465; = &#3627408464;(&#3627408462; − &#3627408463;)+ &#3627408463;(&#3627408464;− &#3627408465;)
Shartga ko’ra &#3627408462; − &#3627408463; > 0,&#3627408464; − &#3627408465; > 0,&#3627408463; > 0,&#3627408464; > 0. Shuning uchun (&#3627408462; − &#3627408463;)+
&#3627408463;(&#3627408464;− &#3627408465;)> 0, ya’ni &#3627408462;&#3627408464; − &#3627408463;&#3627408465; > 0, bundan &#3627408462;&#3627408464; > &#3627408463;&#3627408465; . ▲

§2. O’rtacha qiymatlar va ular orasidagi munosabatlar.
1. O’rtacha qiymatlar.

a ={a 1, a 2 ,…, a n} musbat sonlar ketma-ketligi uchun
o’rta arifmetik qiymat A(a)=A n= naaa
n + + + ... 21 ,
o’rta geometrik qiymat G(a)=G n= n
n aaa
... 21 ,
o’rta kvadratik qiymat K(a)= K n= n aaa
n
2
2
2
2
1 ... +++ va
o’rta garmonik qiymat N (a)=N n= 1
1
2
1
1 ... −
−− +++ naaa
n larni aniqlaymiz.
Xususan x, y musbat sonlar uchun bu o’rta qiymatlar quyidagicha aniqlanadi:

A 2= 2yx
+ ; G 2= xy ; K 2= 2
22yx
+ ; N 2 = yx
xy
+
2.
2. O’rta arifmetik va o’rta geometrik
qiymatlar haqida Koshi tengsizligi va
uning turli isbotlari.
Teorema. A n ≥ G n va An = G n tenglik faqat va faqat
a1=a 2 =…= a n tenglik bo’lganda o’rinli.
1-Isboti . x≥ 1 da ex - 1 ≥ x ekanligi ma’lum, ex -1 =x tenglik esa faqat x= 1 da
bajariladi
. Bundan:
1= e0 = exp ⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛ − ∑=
1
)(
1
n
i i aA
a = )1
)(
exp(
1
− ∏= n
i i aA
a ≥ ∏= n
i i aA
a 1 )( =
n
aA а
G
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
)()( .
Demak, An ≥ G n va tenglik esa faqat 1
() i
na
Aa = , i= 1, 2 ,…, n bulganda bajariladi.
Bundan esa
a1=a 2 =…= a n = A n ekanligi kelib chiqadi .
5

n n A G
≥ ekanligini isbotlaymiz: 2
n = da 12
12
2
aa
aa +
⋅≤
. Bu tengsizlik
ixtiyoriy musbat va sonlar uchun o’rinli bo’lgan
1a 2a ()
2
12
0
aa − ≥
tengsizlikdan oson hosil qilinadi. Berilgan tengsizlikni ixtiyoriy
p ta natural sonlar
uchun to’g’ri deb,
p+1 ta natural sonlar uchun to’g ’riligini isbotlaymiz. Bu sonlar
bo’lsin.
12, , ..., , nn aa a a +1 1
n a + ularning orasida eng kattasi bo’lsin. Ya’ni,
. Shuning uchun
11 1 , ... nnaaa ++ ≥ na
≥ 12
1 ... n
n aa a
a
n
+
+ ++
≥ . Quyidagicha
belgilash kiritamiz:
12 121
1 ... ...
,
11 nn n
nn aa a aa a a nAa
AA
nn 1
n n
n
+ +
+ +++ ++++ ⋅ +
===
+ +
.
bo’lgani uchun
1
n a + ≥ nA 1
nn aA α + = + deb yozish mumkin, bu yyerda 0 α ≥ . U
holda
1 11
nn
n nA A
A
nn
nA α α
+
⋅+ +
==
++ + . Bu tenglikni ikkala qismini ( p+1) –
darajaga ko’tarib, quyidagini topamiz:
() () ()
() () ()( )()
1
11 1
11
1 1
...
11
.
n
nn
nn nnn
nn n n
nn nn nn
AA ACA nn
AA AA AA αα
αα +
++
++
+ +
⎛⎞
=+ = + + ⎜⎟
++
⎝⎠
≥+⋅=⋅+=⋅ n ≥

Farazga ko’ra, ( ) 12 ... n
n
n A aa a
≥⋅⋅⋅ . Buni e’tiborga olib,
() ()
1
11 12 ... nn
nnn n
1n A Aa aa aa +
++ ≥⋅≥⋅⋅⋅⋅ + . Bundan 1
112 ... n
nn 1
n A aa a a + + + ≥⋅⋅⋅⋅ .
Tenglik bo’lganda o’rinli bo’ladi.
12 ... n aa a===

2-isbot . Teoremaning isboti quyidagi tasdiqqa asoslangan:

6

Agar nomanfiy sonlar 12, , ..., n bb b 12 ... 1 n bb b ⋅ ⋅⋅ = tenglikni qanoatlantirsa, u
holda .
12 ... n bb b+++≥ n
Bu tasdiqni masalani matematik induktsiya usulida isbotlaymiz.
1
n = da masala ravshan. nk = da 12 ... 1 k bb b ⋅ ⋅⋅ = tenglikni qanoatlantiruvchi
ixtiyoriy – nomanfiy sonlar uchun
12, , ..., k bb b 12 ... k bb b k + ++ ≥ tengsizlik o’rinli
bo’lsin. da
1
nk =+ 12 1 ... 1 k bb b + ⋅ ⋅⋅ = tenglikni qanoatlantiruvchi ixtiyoriy
– nomanfiy sonlar uchun
12, , ..., k bb b +1 12 1 ... 1 kk bb b b + ⋅ ⋅⋅ ⋅ ≥ tengsizlikni
qanoatlantirishini ko’rsatamiz.
Umumiylikka zarar etkazmasdan 1 1 kbb k+ ≤ ≤ deb hisoblaymiz. Unda
bo’lgani uchun induktsiya faraziga ko’ra
bo’ladi. Endi
() 12 1 1 ... 1 kkk bb b b b −+ ⋅⋅⋅ ⋅ ⋅ =
12 1 1 ... kkk bb b bb k −+ +++ +⋅ ≥ 11 1 kk kkbb bb ++ + ≥⋅ − ekanligini
isbotlash etarli. Bu
() ( ) 1 1 kkbb + +⋅ −≥ 10 tengsizlikka teng kuchli 1 1 kkbb + ≤ ≤
bo’lgani uchun ohirgi tengsizlik o’rinli ekanligi ravshan.

3-isbot . Teoremaning isboti quyidagi ma’lum tasdiqqa asoslangan:
x≥ 1 da ex - 1≥ x , shu bilan birga ex -1 =x tenglik esa faqat x= 1 da bajariladi .
Bundan:
1= e0 = exp ⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛ − ∑=
1
)(
1
n
i i aA
a = )1
)(
exp(
1
− ∏= n
i i aA
a ≥ ∏= n
i i aA
a 1 )( =
n
aA а
G
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
)()( .
Demak, A(a) ≥ G(a) va tenglik esa faqat 1
)( =
aA
a i , i= 1, 2 ,…, n bo’lganda
bajariladi. Bundan esa
a1=a 2 =…= a n = A(a) ekanligi kelib chiqadi .

1-misol . , 0 xy > bo’lsa, 22 1 x yxyx y + +≥ + + tengsizlikni isbotlang.
Yechilishi:
7

$2222 22 22 1 222 2 11 1 22 x xyy xy xyxy xy ++≥++⇒ + + + ++ = + + + . 22 2 22 2 , 22 1 ,1 22 1 . 22 xy xy y yxyxyx x x ⎧ +≥ ⎪ ⎪ ⎪ ++≥ ⇒++≥++ ⎨ ⎪ ⎪ +≥ ⎪ ⎩ . y 2-misol . 0 x > bo’lsa, 6 12 4 2222 x x x + ≥⋅ tengsizlikni isbotlang. Yechilishi. 1 11 6 12 4 12 4 6 12 4 2222222 2222 x xxxx x + + ≥⋅ ⋅ =⋅ =⋅ =⋅ x 0 . Misollar: 1. Agar ,xy > bo’lsa, 44 88 x y ++≥ xy 0 ni isbotlang. 2. bo’lsa quyidagini isbotlang: 12 345,,,, xx xxx > ( ) 22222 1 23 45 123 45x xxxx xxxxx ++++≥ +++ . 3. ,, 0x yz > bo’lsa, 222x y z xy yz xz ++≥++ ni isbotlang. 4. bo’lsa, ,, 0abc> 3 abc bca ++ ≥ ni isbotlang. 5. bo’lsa, () ,, 0abc> ( )( )( ) 11 16 abcabc a +++ +≥ bc ni isbotlang. 3. O’rta geometrik va o’rta garmonik qiymatlar orasidagi tengsizli\ k. Teorema. G (a) ≥ H(a) ekanligini, jumladan, H(a) = G(a) tenglik faqat va faqat a1=a 2 =…= a n shart bajarilsa to’g’riligini isbotlang. Isboti . Koshi tengsizligidan foydalanib (1-masalaga qarang) foydalanib 8$