BA’ZI AJOYIB TENGSIZLIKLAR VA ULARNI ISBOTLASH

Аннотация

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 ,𝑥 ,...𝑥 haqiqiy musbat sonlar bo’lib, 1 2 𝑛 1 1 1 1 + +...+ = 𝑥 +1998 𝑥 + 1998 𝑥 + 1998 𝑥 + 1998 1 2 𝑛 1 tenglikni qanoatlantirsin, 𝑛√𝑥 𝑥 ...𝑥 1 2 𝑛 ≥ 1998 𝑛 − 1 tengsizlikni isbotlang. 1 1 1 Yechilishi. 𝑦 = bo’lsin. U holda 𝑦 +...+𝑦 = va 𝑥 = − 1998 𝑖 1 𝑛 𝑖 𝑥 +1998 1998 𝑦 1 𝑖 bo’ladi. Quyidagi tenglik o’rinli: 𝑛 𝑛 1 ∑𝑛 𝑙𝑛( 1 −1998) ∏𝑥 = ∏( − 1998) = 𝑒 𝑖=1 𝑦 𝑖 𝑖 𝑦 𝑖 𝑖=1 𝑖=1 1 Demak, 𝑥 larning ko’paytmasini mimimallashtirish, 𝑙𝑛( − 1998) lar yig’indisini 𝑖 𝑦 𝑖 minimallashtirishga ekvivalent. Bundan https://scientific-jl.com/new 33 99 Volume–75_Issue-1_April-2025 JOURNAL OF NEW CENTURY INNOVATIONS 𝑑 1 1 −1 −1 (𝑙𝑛( − 1998)) = ∙ = 𝑑𝑦 𝑦 1 𝑦2 𝑦 − 1998𝑦2 ( −1998)2 𝑦 𝑑2 1 1 − 3996𝑦 (𝑙𝑛( − 1998)) = 𝑑𝑦2 𝑦 (𝑦 − 1998𝑦2)2 tenglik o’rinli ekan. 1 1 Demak, 𝑙𝑛( − 1998) funksiya [0; ] da qavariq ekan. Agar barcha i lar uchun 0 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 + 𝑐 √𝑏 + 𝑎 4(𝑎 + 𝑏 + 𝑐) + + ≥ 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 4(𝑎 + 𝑏 + 𝑐) 𝑎 √(𝑎 + 𝑏)(𝑏 + 𝑐)(𝑐 + 𝑎) 𝑐𝑦𝑐 𝑏 + 𝑐 ↔ ∑ √(𝑎 +𝑏)(𝑐 + 𝑎) ≥ 4(𝑎 + 𝑏 + 𝑐) 𝑎 𝑐𝑦𝑐 Koshi tengsizligidan kelib chiqadigan natijani almashtirami