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39
BA’ZI AJOYIB TENGSIZLIKLAR VA ULARNI ISBOTLASH
Jumayeva Dilrabo Xolmurot qizi
Qorako’l tuman 1-son politexnikum o’qituvchisi
E-mail: jumayevadilrabo414@gmail.com
Annotatsiya:
Maqolada musbat hamda ma’lum bir berilgan shartlarni
qanoatlantiradigan sonlar uchun tengsizliklar hamda ularning isboti berilgan. Isbotlash
maqsadida Koshi hamda Bernulli tengsizliklaridan foydalanilgan.
Kalit so’zlar
: Koshi tengsizligi, musbat son, Chebishev tengsizligi, qavariq
funksiya, Bernulli tengsizligi
Annotation:
The article gives inequalities as well as proofs for numbers
satisfying positive and certain given conditions. For proof purposes, Koshi as well as
Bernoulli inequalities have been used.
Key words:
Koshi inequality, positive number, Chebyshev inequality, convex
function, Bernoulli inequality.
Musbat
haqiqiy
sonlar
ustida
berilgan
shartlar
asosida
berilgan
tengsizliklarning o’rinli ekanligini isbotlashning turlicha usullari mavjud ular ulardan
eng ko’p qo’llaniladigan sonlarning o’rta arifmetigi hamda o’rta geometrigi haqidagi
Koshi tengsizligidir. Bundan tashqari tengsizlikni kattalashtirib baholash yoki
Chebishev tengsizligidan foydalanish kabi usullar ham mavjud. Bularning qaysi
biridan foydalanish mutlaqo o’zimizga bog’liq. Quyida bir nechta tengsizliklar va
ulaning isbotini ko’rib chiqamiz:
1-misol.
Aytaylik,
𝑥
1
, 𝑥
2
, . . . 𝑥
𝑛
haqiqiy musbat sonlar bo’lib,
1
𝑥
1
+ 1998
+
1
𝑥
2
+ 1998
+. . . +
1
𝑥
𝑛
+ 1998
=
1
𝑥
1
+ 1998
tenglikni qanoatlantirsin,
√𝑥
1
𝑥
2
. . . 𝑥
𝑛
𝑛
𝑛 − 1
≥ 1998
tengsizlikni isbotlang.
Yechilishi.
𝑦
𝑖
=
1
𝑥
1
+1998
bo’lsin. U holda
𝑦
1
+. . . +𝑦
𝑛
=
1
1998
va
𝑥
𝑖
=
1
𝑦
𝑖
− 1998
bo’ladi. Quyidagi tenglik o’rinli:
∏ 𝑥
𝑖
𝑛
𝑖=1
= ∏(
𝑛
𝑖=1
1
𝑦
𝑖
− 1998) = 𝑒
∑
𝑙𝑛(
1
𝑦
𝑖
−1998)
𝑛
𝑖=1
Demak,
𝑥
𝑖
larning ko’paytmasini mimimallashtirish,
𝑙𝑛(
1
𝑦
𝑖
− 1998)
lar yig’indisini
minimallashtirishga ekvivalent. Bundan
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𝑑
𝑑𝑦
(𝑙𝑛(
1
𝑦
− 1998)) =
1
(
1
𝑦 − 1998)
2
∙
−1
𝑦
2
=
−1
𝑦 − 1998𝑦
2
𝑑
2
𝑑𝑦
2
(𝑙𝑛(
1
𝑦
− 1998)) =
1 − 3996𝑦
(𝑦 − 1998𝑦
2
)
2
tenglik o’rinli ekan.
Demak,
𝑙𝑛(
1
𝑦
− 1998)
funksiya
[0;
1
3996
]
da qavariq ekan. Agar barcha
i
lar uchun
0 <
𝑦
𝑖
<
1
3996
bo’lsa biz Jensen almashtirishidan foydalanishimiz mumkin edi. Demak,
𝑦
𝑖
+ 𝑦
𝑗
≤
1
1998
shartni qanoatlantiruvchi barcha i, j lar uchun tasodifiy olingan
ixtiyoriy
a + b ≤
1
1998
uchun amal qiladigan quyidagi tengsizlikni ko’rib
chiqamiz:
(
1
a
− 1998)(
1
b
− 1998) ≥ (
2
a + b
− 1998)
2
↔
1
ab
− 1998(
1
a
+
1
b
) ≥
4
(a + b)
2
−
4 ∙ 1998
(a + b)
↔ (a + b)
2
− 1998(a + b)
3
≥ 4ab − 4ab(a + b) ∙ 1998
↔ (a − b)
2
≥ 1998(a + b)(a − b)
2
Shu bilan birga qaralayotgan yig’indini kamaytirish uchun har qanday ikkita
𝑦
𝑖
va
𝑦
𝑗
larning o’rta qiymatini mos qo’yish mumkin. Shuning uchun
𝑦
𝑖
∈ (0;
1
3996
]
deb faraz
qilamiz. Jensen tengsizligidan barcha
i
lar uchun
𝑦
𝑖
=
1
1998𝑛
bo’lganda yoki barcha
i
lar uchun
𝑥
𝑖
= 1998(𝑛 − 1)
bo’lganda minimum qiymatga erishadi.
Bundan esa tengsizlik osongina kelib chiqishi ko’rinib turibdi.
2-misol.
𝑥
1
𝑥
2
. . . 𝑥
𝑛
= 1
tenglik bilan aniqlangan barcha musbat haqiqiy
𝑥
1
, 𝑥
2
, . . . 𝑥
𝑛
lar uchun
1
𝑛 − 1 + 𝑥
1
+. . . +
1
𝑛 − 1 + 𝑥
𝑛
≤ 1
o’rinli bo’lishini ko’rsating.
Yechilishi:
Dastlab quyidagi lemmani isbotlaymiz: Yig’indining maximal qiymati
𝑥
𝑖
n − 1
ga teng bo’lganda hosil bo’ladi. Ixtiyoriy nomanfiy k o’zgarmas son uchun
f(y) =
1
k+e
y
ni ko’rib chiqaylik. Bizda
𝑓
′
(𝑦) =
𝑒
−𝑦
(𝑘+𝑒
𝑦
)
2
va
𝑓
′′
(𝑦) =
𝑒
𝑦
(𝑒
𝑦
−𝑘)
(𝑘+𝑒
𝑦
)
3
o’rinli ekanligi bor.
𝑓
′′
(𝑦) ≥ 0 ↔ 𝑒
𝑦
≥ 𝑘
o’rinli. Demak,
f(y)
funksiya
(ln(k); ∞)
da qavariq bo’ladigan
y = lnk
qiymatida
f(y)
funksiya yagona egilish
nuqtasiga ega. Endi
𝑦
1
+. . . +𝑦
𝑛
= 0
va
∑
1
𝑛−1+𝑥
𝑖
= ∑
𝑓(𝑦
𝑖
)
𝑛
𝑖=1
𝑛
𝑖=1
tengliklarni qanoatlantiradigan
𝑦
𝑖
=
𝑙𝑛(𝑥
𝑖
)
ketma ketlikni kiritamiz. Bundan
k = n − 1
va
𝑘
0
= 𝑙𝑛(𝑛 − 1)
deb yozish
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mumkin. Ba’zi musbat
m
lar uchun
𝑦
1
≥. . . ≥ 𝑦
𝑚
≥ 𝑘
0
≥ 𝑦
𝑚+1
≥. . . 𝑥
𝑛
o’rinli
deylik. Bundan quyidagi kattalashtirish o’rinli:
f(𝑦
1
)+. . . +𝑓(𝑦
𝑚
) ≤ (𝑚 − 1)𝑓(𝑘
0
) + 𝑓(𝑦
1
+. . . +𝑦
𝑚
− (𝑚 − 1)𝑘
0
)
Shuningdek,
(𝑚 − 1)𝑓(𝑘
0
) + 𝑓(𝑦
𝑚+1
)+. . . +𝑓(𝑦
𝑛
) ≤ (𝑛 −
1)𝑓(
(𝑚−1)𝑘
0
+𝑦
𝑚+1
+...+𝑦
𝑛
)
𝑛−1
) kattalashtirish ham o’rinli boshqa tomondan barcha
𝑦
𝑖
lar
𝑘
0
lardan kichik. Shu sababli biz
𝑦
𝑖
ni
𝑛 − 1
ga tenglashtirish uchun
kattalashtirishni qo’llaymiz, shu bilan birga ko’rib chiqilayotgan yig’indini
oshiramiz. Demak, lemma isbotlandi.
Lemmani qo’llab
𝑘
𝑘+𝑥
+
𝑘
𝑘+
1
𝑥𝑘
≤ 1
tengsizlikning to’g’riligini ko’rsatishimiz yetarli.
Maxrajlardan qutulamiz,
(𝑘
2
+
𝑘
𝑥
𝑘
) + (𝑘 + 𝑥) ≤ 𝑘
2
+ 𝑘 (𝑥 +
1
𝑥
𝑘
) + 𝑥
1−𝑘
− 𝑥𝑘 + 𝑥 + 𝑘 ≤ 𝑥
1−𝑘
Ammo bu aniq edi. Bernulli tengsizligidan
𝑥
1−𝑘
= (1 + (𝑥 − 1)
1−𝑘
> 1 + (𝑥 − 1)(1 − 𝑘) = 𝑥 + 𝑘 − 𝑥𝑘
o’rinli.
Bundagi
tenglik faqat
x = 1
yoki
n = 2
da bajariladi.
3-misol.
Barcha musbat haqiqiy
a, b, c
sonlari uchun
√𝑏 + 𝑐
𝑎
+
√𝑎 + 𝑐
𝑏
+
√𝑏 + 𝑎
𝑐
≥
4(𝑎 + 𝑏 + 𝑐)
√(𝑎 + 𝑏)(𝑏 + 𝑐)(𝑐 + 𝑎)
tengsizlik o’rinli bo’lishini isbotlang.
Yechilishi.
1-usul
. Koshi tengsizligiga ko’ra,
√(𝑎 + 𝑏)(𝑐 + 𝑎) ≥ 𝑎 + √𝑏𝑐.
o’rinli bo’lishi
aniq . Bundan quyidagini yozish mumkin:
∑
√𝑏 + 𝑐
𝑎
≥
𝑐𝑦𝑐
4(𝑎 + 𝑏 + 𝑐)
√(𝑎 + 𝑏)(𝑏 + 𝑐)(𝑐 + 𝑎)
↔ ∑
𝑏 + 𝑐
𝑎
√(𝑎 + 𝑏)(𝑐 + 𝑎) ≥ 4(𝑎 + 𝑏 + 𝑐)
𝑐𝑦𝑐
Koshi tengsizligidan kelib chiqadigan natijani almashtiramiz va
∑
(𝑏 + 𝑐)
√𝑏𝑐
𝑎
≥
𝑐𝑦𝑐
2(𝑎 + 𝑏 + 𝑐)
tengsizlikni ko’rsatishimiz yetarli.
a > b > c
ekanligidan
b + c ≤ c + a ≤ a + b va
√𝑏𝑐
𝑎
≤
√𝑎𝑐
𝑏
≤
√𝑏𝑎
𝑐
o’rinli. AM-
GM va Chebishev tengsizligidan
2-usul.
Aytaylik,
x = √𝑏 + 𝑐
,
y = √𝑎 + 𝑐
,
z = √𝑏 + 𝑎
bo’lsin.
x, y, z
sonlar
XYZ
o’tkir burchakli uchburchakning tomonlari deb olsak,
𝑥
2
+ 𝑦
2
= 𝑎 + 𝑏 +
2𝑐 > 𝑎 + 𝑏 = 𝑧
2
o’rinli. Yuqoridagi tengsizlik esa
∑
𝑥
𝑦
2
+ 𝑧
2
− 𝑥
2
≥
𝑥
2
+ 𝑦
2
+ 𝑧
2
𝑥𝑦𝑧
𝑐𝑦𝑐
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Tengsizlikka ekvivalentdir.
𝑦
2
+ 𝑧
2
− 𝑥
2
= 2𝑦𝑧𝑐𝑜𝑠(𝑋)
ifoda o’rinli ekanligidan
bu ekvivalent ifodani kamaytiramiz:
∑
𝑥
2
𝑐𝑜𝑠(𝑋)
≥ 2(𝑥
2
+ 𝑦
2
+ 𝑧
2
)
𝑐𝑦𝑐
x ≥ y ≥ z
ekanligidan
1
𝑐𝑜𝑠(𝑋)
≥
1
𝑐𝑜𝑠(𝑌)
≥
1
𝑐𝑜𝑠(𝑍)
o’rinli natijada chap tomonga
Chebishev tengsizligini qo’llasak, kosinuslar o’zaro yig’indisining kamida 6
ekanligini isbotlash uchun kerakli miqdorni kamaytiradi, ya’ni AM-HM dan
1
𝑐𝑜𝑠(𝑋)
+
1
𝑐𝑜𝑠(𝑌)
+
1
𝑐𝑜𝑠(𝑍)
≥
9
𝑐𝑜𝑠(𝑋)+𝑐𝑜𝑠(𝑌)+𝑐𝑜𝑠(𝑍)
o’rinli . Ammo geometriyada
Uchburchak qoidalarini eslasak,
𝑐𝑜𝑠(𝑋) + 𝑐𝑜𝑠(𝑌) + 𝑐𝑜𝑠(𝑍) = 1 +
𝑟
𝑅
va
R ≥ 2r
tengsizlik bajarilar edi, demak faraz qilgan holatimiz isbotlandi.
4-misol.
Barcha musbat haqiqiy
𝑥
1
, 𝑥
2
, . . . 𝑥
𝑛
uchun
𝑥
1
3
𝑥
1
2
+𝑥
1
𝑥
2
+𝑥
2
2
+
𝑥
2
3
𝑥
2
2
+𝑥
2
𝑥
3
+𝑥
3
2
+. . . +
𝑥
𝑛
3
𝑥
𝑛
2
+𝑥
𝑛
𝑥
1
+𝑥
1
2
≥
𝑥
1
+...+𝑥
𝑛
3
tengsizlikni isbotlang.
Yechilishi.
Bizga yaxshi ma’lumki,
0 = (𝑥
1
-
𝑥
2
) + (𝑥
2
− 𝑥
3
)+. . . +(𝑥
𝑛
− 𝑥
1
) = ∑
𝑥
𝑖
3
−𝑥
𝑖+1
3
𝑥
𝑖
2
+𝑥
𝑖
𝑥
𝑖+1
+𝑥
𝑖+1
2
𝑛
𝑖=1
tenglik o’rinli
(bu yerda
𝑥
𝑛+1
= 𝑥
1
).
Shuning uchun
∑
𝑥
𝑖
3
𝑥
𝑖
2
+𝑥
𝑖
𝑥
𝑖+1
+𝑥
𝑖+1
2
𝑛
𝑖=1
=
1
2
∑
𝑥
𝑖
3
+𝑥
𝑖+1
3
𝑥
𝑖
2
+𝑥
𝑖
𝑥
𝑖+1
+𝑥
𝑖+1
2
𝑛
𝑖=1
tenglik bajariladi.
Ammo
𝑎
3
+ 𝑏
3
≥
1
3
𝑎
3
+
2
3
𝑎
2
𝑏 +
2
3
𝑎𝑏
2
+
1
3
𝑏
3
=
1
3
(𝑎 + 𝑏)(𝑎
2
+ 𝑎𝑏 + 𝑏
2
)
tenglikdan
1
2
∑
𝑥
𝑖
3
+𝑥
𝑖+1
3
𝑥
𝑖
2
+𝑥
𝑖
𝑥
𝑖+1
+𝑥
𝑖+1
2
𝑛
𝑖=1
≥
1
2
∑
𝑥
𝑖
+𝑥
𝑖+3
3
=
1
3
∑
𝑥
𝑖
𝑛
𝑖=1
𝑛
𝑖=1
o’rinli ekanligini aytish
mumkin.
5-misol.
Aytaylik,
a, b, c
musbat haqiqiy sonlar bo’lib,
a + b ≥ c;
b + c ≥ a; c + a ≥ b
bo’lsin. U holda
2𝑎
2
(𝑏 + 𝑐) + 2𝑏
2
(𝑎 + 𝑐) + 2𝑐
2
(𝑏 + 𝑎) ≥ 𝑎
3
+ 𝑏
3
+ 𝑐
3
+ 9𝑎𝑏𝑐
tengsizlik bajarilishini ko’rsating.
Yechilishi.
Tengsizlikni
isbotlash
uchun
a = y + z, b = z + x, c = x + y
almashtirishni bajaramiz. Chap tomondagi ifoda
4𝑥
3
+ 4𝑦
3
+ 4𝑧
3
+ 10𝑥
2
(𝑦 + 𝑧) + 10𝑦
2
(𝑧 + 𝑥) + 10𝑧
2
(𝑥 + 𝑦) + 24𝑥𝑦𝑧
ko’rinishga, o’ng tomondagi ifoda esa
2𝑥
3
+ 2𝑦
3
+ 2𝑧
3
+ 12𝑥
2
(𝑦 + 𝑧) + 12𝑦
2
(𝑧 + 𝑥) + 12𝑧
2
(𝑥 + 𝑦) + 18𝑥𝑦𝑧
ko’rinishga keladi. Bu ifoda Shur tengsizligi deb ataluvchi
𝑥
3
+ 𝑦
3
+ 𝑧
3
+ 3𝑥𝑦𝑧 ≥ 𝑥
2
(𝑦 + 𝑧) + 𝑦
2
(𝑧 + 𝑥) + 𝑧
2
(𝑥 + 𝑦)
tengsizlikka ekvivalentdir.
Tenglik
x = y = z
bo’lganda ya’ni
(a, b, c) = (t, t, t)
yoki
x, y, z
larni ikkitasi teng,
uchinchisi nol bo’lganda ya’ni
(a, b, c)ϵ{(2t, t, t), (t, 2t, t), (t, t, 2t)}
bo’lganda bajariladi.
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6-misol.
a, b, c
sonlari uchburchak tomonlari bo’lsa,
𝑎
√2𝑏
2
+ 2𝑐
2
− 𝑎
2
+
𝑏
√2𝑎
2
+ 2𝑐
2
− 𝑏
2
+
𝑐
√2𝑏
2
+ 2𝑎
2
− 𝑐
2
≥ √3
Yechilishi.
Dastlab
a = y + z, b = z + x, c = x + y
almashtirish bajaramiz, bizga
yaxshi ma’lumki
x, y, z
musbat sonlar. Demak,
∑
𝑎
√2𝑏
2
+ 2𝑐
2
− 𝑎
2
= ∑
𝑦 + 𝑧
√4𝑥
2
+ 4𝑥𝑦 + 4𝑥𝑧 + 𝑦
2
+ 𝑧
2
− 2𝑦𝑧
𝑐𝑦𝑐
𝑐𝑦𝑐
tenglik o’rinli.
f(x) =
1
√x
qavariq funksiyani ko’rib chiqaylik. (Ko’rinib turibdiki, Jusen har doim
tengsizliklardan radikallarni yo’q qilish uchun qulay almashtirishni taqdim qiladi.)
x +
y + z = 1
bo’lsin, u holda
∑(𝑦 + 𝑧)𝑓(
𝑐𝑦𝑐
4𝑥
2
+ 4𝑥𝑦 + 4𝑥𝑧 + 𝑦
2
+ 𝑧
2
− 2𝑦𝑧) ≥
((𝑦 + 𝑧) + (𝑧 + 𝑥) + (𝑥
+ 𝑦))𝑓 (
∑
(𝑦 + 𝑧)(4𝑥
2
+ 4𝑥𝑦 + 4𝑥𝑧 + 𝑦
2
+ 𝑧
2
− 2𝑦𝑧)
𝑐𝑦𝑐
(𝑦 + 𝑧) + (𝑧 + 𝑥) + (𝑥 + 𝑦)
)
=
2√2
√∑
4𝑥
2
(𝑦 + 𝑧) + (4𝑥𝑦
2
+ 4𝑥𝑦𝑧) + (4𝑥𝑦𝑧 + 4𝑥𝑧
2
)
𝑐𝑦𝑐
+ 𝑦
3
+ 𝑧
3
− 𝑦
2
𝑧 − 𝑦𝑧
2
∑ 4𝑥
2
(𝑦 + 𝑧) + (4𝑥𝑦
2
+ 4𝑥𝑦𝑧) + (4𝑥𝑦𝑧 + 4𝑥𝑧
2
)
𝑐𝑦𝑐
+ 𝑦
3
+ 𝑧
3
− 𝑦
2
𝑧 − 𝑦𝑧
2
= ∑ 2𝑥
3
+ 7𝑥
2
(𝑦 + 𝑧) + 8𝑥𝑦𝑧,
𝑐𝑦𝑐
8(𝑥 + 𝑦 + 𝑧)
3
≥ 3 ∑ 2𝑥
3
+ 7𝑥
2
(𝑦 + 𝑧) + 8𝑥𝑦𝑧,
𝑐𝑦𝑐
↔ ∑ 4𝑥
3
+ 24𝑥
2
𝑦 + 8𝑥𝑦𝑧 ≥ ∑ 3𝑥
3
+ 21𝑥
2
𝑦 + 12𝑥𝑦𝑧 ≥
𝑠𝑦𝑚
𝑠𝑦𝑚
↔ 2𝑥
3
+ 2𝑦
3
+ 2𝑧
3
+ 3(𝑥
2
(𝑦 + 𝑧) + 𝑦
2
(𝑥 + 𝑧) + 𝑧
2
(𝑦 + 𝑥)) ≥ 24𝑥𝑦𝑧
Adabiyotlar ro’yxati:
1.
Hardy, G. H., Littlewood, J. E., & Pólya, G. Inequalities. Cambridge University
Press, 1934.
2. Mitrinović, D. S. Analytic Inequalities. Springer-Verlag, 1970.
3. Mitrinović, D. S., Pečarić, J. E., & Fink, A. M. Classical and New Inequalities
in Analysis. Kluwer Academic Publishers, 1993.
4. Beckenbach, E. F., & Bellman, R. Inequalities. Springer-Verlag, 1961.
5. Bullen, P. S. Handbook of Means and Their Inequalities. Springer, 2003.
JOURNAL OF NEW CENTURY INNOVATIONS
Volume–75_Issue-1_April-2025
44
44
6. Steele, J. M. The Cauchy-Schwarz Master Class: An Introduction to the Art of
Mathematical Inequalities. Cambridge University Press, 2004.
