GERRETSEN TENGSIZLIGI VA UNING TADBIQLARIGA DOIR BA'ZI MASALALAR TAHLILI

Educational Research in Universal Sciences ISSN: 2181-3515 VOLUME 4 | SPECIAL ISSUE 12 | 2025 https://t.me/Erus_uz Multidisciplinary Scientific Journal October, 2025 131 DOI: https://10.5281/zenodo.17372283 GERRETSEN TENGSIZLIGI VA UNING TADBIQLARIGA DOIR BA’ZI MASALALAR TAHLILI Shukurullayeva Vasila Fazliddinovna E-mail: [email protected] Buxoro davlat universiteti, Buxoro, O‘zbekiston Shukurullayeva Mohinur Fazliddinovna E-mail: [email protected] Buxoro davlat universiteti, Buxoro, O‘zbekiston ANNOTATSIYA Ushbu ishda Gerretsen teoremasi qaralgan. Ularga doir o‘quvchilar hamda talabalar o‘rtasida bo‘lib o‘tadigan fan olimpiadalaridagi ba’zi masalalar yechimlari tahlil qilingan. Mustaqil ishlash uchun masalalar keltirilgan. Kalit so‘zlar. Gerretsen teoremasi, uchburchak, uchburchakka ichki va tashqi chizilgan aylana radiusi, yarim perimetr. So‘nggi yillarda geometriyada tengsizliklar usullari alohida o‘rin egallab, olimpiada va oliy ta’lim kurslarida eng samarali vositalardan biriga aylandi. Uchburchak va ko‘pburchaklarga doir tengsizliklarni yechishda ko‘pincha bitta formulani emas, balki ketma-ket qo‘llanadigan umumiy usullarni bilish hal qiluvchi ahamiyatga ega. Bu o‘rinda keng qo‘llaniladigan tengsizliklardan biri Gerretsen tengsizligidir. Gerretsen tengsizligi uchburchakning unga ichki va tashqi chizilgan aylana radiusi hamda yarim perimetri orasidagi bog‘lanishni ifodalaydi. Teorema.Ixtiyoriy ∆𝐴𝐵𝐶 uchburchak uchun quyidagi tengsizlik bajariladi: (Bu yerda R-tashqi chizilgan aylana radiusi, richki chizilgan aylana radiusi, 𝑠= 𝑎+𝑏+𝑐 2) 16𝑅𝑟−5𝑟2≤𝑠2≤4𝑅2+4𝑅𝑟+3𝑟2 1-misol. Ixtiyoriy ∆𝐴𝐵𝐶 uchburchak uchun quyidagi tengsizlikni isbotlang: 𝑠2≤4𝑅2+8𝑅𝑟 Yechimi: Gerretsen teoremasiga ko‘ra: 16𝑅𝑟−5𝑟2≤𝑠2≤4𝑅2+4𝑅𝑟+3𝑟2 Educational Research in Universal Sciences ISSN: 2181-3515 VOLUME 4 | SPECIAL ISSUE 12 | 2025 https://t.me/Erus_uz Multidisciplinary Scientific Journal October, 2025 132 Bizga kerak bo‘lgan maqsad: 𝑠2≤4𝑅2+8𝑅𝑟 4𝑅2+4𝑅𝑟+3𝑟2≤4𝑅2+8𝑅𝑟 3𝑟2≤4 𝑟≤4 3𝑅 Uchburchak uchun Euler tengsizligiga ko‘ra 𝑅≥2𝑟 dan 𝑟≤𝑅 2<4 3𝑅 Gerretson tengsizligidan 𝑠2≤4𝑅2+4𝑅𝑟+3𝑟2≤4𝑅2+8𝑅𝑟 𝑠2≤4𝑅2+8𝑅𝑟 2-misol. Ixtiyoriy ∆𝐴𝐵𝐶 uchburchak uchun quyidagi tengsizlikni isbotlang: 𝑅2−𝑅𝑟+𝑟2≥𝑠2 12 Yechim: 16𝑅𝑟−5𝑟2≤ 𝑠2≤4𝑅2+4𝑅𝑟+3𝑟2 (Gerretson teoremasi) 𝑠2≤12(𝑅2−𝑅𝑟+𝑟2) Ikki tenglikni o‘ng tomonini taqqoslasak: 12(𝑅2−𝑅𝑟+𝑟2)−(4𝑅2+4𝑅𝑟+3𝑟2)=8𝑅2−16𝑅𝑟+9𝑟2 8𝑅2−16𝑅𝑟+9𝑟2=8(𝑅−𝑟)2+𝑟2≥0 4𝑅2+4𝑅𝑟+3𝑟2≤12(𝑅2−𝑅𝑟+𝑟2) 𝑠2≤4𝑅2+4𝑅𝑟+3𝑟2≤12(𝑅2−𝑅𝑟+𝑟2) 𝑠2≤12(𝑅2−𝑅𝑟+𝑟2) → 𝑅2−𝑅𝑟+𝑟2≥𝑠2 12 3-misol. Uchburchak ABC uchun (𝑎𝑏+𝑏𝑐+𝑐𝑎)(𝑠2+𝑟2)≥4𝑎𝑏𝑐 (𝑠+3𝑟) ni isbotlang. Yechim:Bizga malumki, ab+bc+ca=s2+r2+4Rr va abc = 4Rrs Bundan (s2+r2+4Rr)(𝑠2+𝑟2)≥16𝑅𝑟𝑠2+36𝑅2𝑟2 Chap tarafni X=s2+r2deb belgilab ochamiz: 𝑋(𝑋+4𝑅𝑟)=𝑋2+4𝑅𝑟𝑋 𝑠4+2𝑠2𝑟2+𝑟4+4𝑅𝑟𝑠2+4𝑅𝑟3≥16𝑅𝑟𝑠2+36𝑅2𝑟2 Hamma narsani o‘ng tomonga o‘tkazib, 𝑠2-li hadlarni guruhlab yozsak: 𝑠4+𝑠2(2𝑟2−12𝑅𝑟)≥36𝑅2𝑟2−4𝑅𝑟3−𝑟4 (𝑠2−6𝑅𝑟+𝑟2)2=𝑠4+𝑠2(2𝑟2−12𝑅𝑟)+(6𝑅𝑟−𝑟2)2 (𝑠2−6𝑅𝑟+𝑟2)2≥(6𝑅𝑟−𝑟2)2+36𝑅2𝑟2−4𝑅𝑟3−𝑟4 O‘ng tomonni soddalashtirsak: (6𝑅𝑟−𝑟2)2+36𝑅2𝑟2−4𝑅𝑟3−𝑟4= 72𝑅2𝑟2−16𝑅𝑟3 Natijada (𝑠2−6𝑅𝑟+𝑟2)2≥72𝑅2𝑟2−16𝑅𝑟3 Gerretsen tengsizligining quyi bahosi: 𝑠2≥16𝑅𝑟−5𝑟2 Shundan 𝑠2−6𝑅𝑟+𝑟2≥(16𝑅𝑟−5𝑟2)−6𝑅𝑟+𝑟2=10𝑅𝑟−4𝑟2 Educational Research in Universal Sciences ISSN: 2181-3515 VOLUME 4 | SPECIAL ISSUE 12 | 2025 https://t.me/Erus_uz Multidisciplinary Scientific Journal October, 2025 133 ya’ni (𝑠2−6𝑅𝑟+𝑟2)2≥(10𝑅𝑟−4𝑟2)2 Shu bois yuqoridagi natijani tugatish uchun quyidagini tekshirish kifoya: (10𝑅𝑟−4𝑟2)2≥72𝑅2𝑟2−16𝑅𝑟3 ⇔ (𝑅−2𝑟)(7𝑅−2𝑟)≥0 Eyler tengsizligiga ko‘ra 𝑅 ≥ 2 𝑟 dan (𝑅−2𝑟)(7𝑅−2𝑟)≥0 kelib chiqdi ,demak dastlabki tengsizlik to‘liq isbotlandi. (𝑎𝑏+𝑏𝑐+𝑐𝑎)(𝑠2+𝑟2)=4𝑎𝑏𝑐 (𝑠+3𝑟) faqat teng tomonli uchburchakda yuz beradi. 4-misol. Uchburchak ABC uchun isbotlang (ichki markaz I, tashqi aylana radiusi R, ichki radius r) √12(𝑅2−𝑅𝑟+𝑟2)≥𝐴𝐼+𝐵𝐼+𝐶𝐼≥6𝑟 Chap tomon. 𝐴𝐼2=𝑏𝑐−4𝑅𝑟; 𝐵𝐼2=𝑐𝑎−4𝑅𝑟; 𝐶𝐼2=𝑎𝑏−4𝑅𝑟: Koshi-Shvars tengsizligidan: 3(𝐴𝐼2+𝐵𝐼2+𝐶𝐼2)≥(𝐴𝐼+𝐵𝐼+𝐶𝐼)2 ⇔ √3(𝑠2+𝑟2−8𝑅𝑟) ≥ 𝐴𝐼 + 𝐵𝐼 + 𝐶𝐼 Endi quyidagini ko‘rsatamiz: √3(𝑠2+𝑟2−8𝑅𝑟) ≤ √12(𝑅2−𝑅𝑟+𝑟2) bu esa ekvivalent: 𝑠2+𝑟2+8𝑅𝑟≤4𝑅2−4𝑅𝑟+4𝑟2 ⇔ 𝑠2≤4𝑅2+4𝑅𝑟+ 3𝑟2 O‘ng tomon. Ma’lumki: 𝐴𝐼 = 𝑟 𝑠𝑖𝑛𝐴 2 𝐵𝐼=𝑟 𝑠𝑖𝑛𝐵 2 𝐶𝐼=𝑟 𝑠𝑖𝑛𝐶 2 6 𝑟 ≤ 𝑟 𝑠𝑖𝑛𝐴 2 + 𝑟 𝑠𝑖𝑛𝐵 2 + 𝑟 𝑠𝑖𝑛𝐶 2 ya’ni 2 ≤ 1 𝑠𝑖𝑛𝐴 2+1 𝑠𝑖𝑛𝐵 2+1 𝑠𝑖𝑛𝐶 2 3 1 𝑠𝑖𝑛𝐴 2 + 1 𝑠𝑖𝑛𝐵 2 + 1 𝑠𝑖𝑛𝐶 2 3 ≥ 𝑐𝑠𝑐𝐴+𝐵+𝐶 6 A + B + C =π bo‘lganidan, 𝑐𝑠𝑐(𝜋 /6) = 2 va natija keladi: AI + BI + CI ≥ 6 r. 5-misol. Ixtiyoriy ∆𝐴𝐵𝐶 uchburchak uchun quyidagi tengsizlikni isbotlang: ∑1 2 − 𝑐𝑜𝑠𝐴≥2≥3∑1 5 − 𝑐𝑜𝑠𝐴 ∑cosA=1+r/R, ∑cosAcosB=(s2+r2−4R2)/(4R2) , Educational Research in Universal Sciences ISSN: 2181-3515 VOLUME 4 | SPECIAL ISSUE 12 | 2025 https://t.me/Erus_uz Multidisciplinary Scientific Journal October, 2025 134 ∏cosA=(s2−(2R+r)2)/(4R2) Chap tomoni: ∑1/(2−cosA)≥2 Eyler 2𝑟 ≤ 𝑅 va Gerretsen s2≤4R2+4Rr+3r2 dan: 𝑠2≤4𝑅2+8𝑅𝑟−5𝑟2 Identitetlar orqali quyidagiga teng kuchli: 4(1+𝑟/𝑅)+2⋅𝑠2−(2𝑅+𝑟)2 4𝑅2≥4+3⋅𝑠2+𝑟2−4𝑅2 4𝑅2 → 4∑𝑐𝑜𝑠𝐴+2∏𝑐𝑜𝑠𝐴≥4+3∑𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 (2 − 𝑐𝑜𝑠 𝐴) bilan ekvivalent ko‘rinish: ∑(2−𝑐𝑜𝑠𝐴)(2−𝑐𝑜𝑠𝐵)≥2⋅∏(2−𝑐𝑜𝑠𝐴) Shundan: ∑1 2 − 𝑐𝑜𝑠𝐴≥2 O‘ng tomoni: 3∑1 5 − 𝑐𝑜𝑠𝐴≤2 Eyler 2𝑟 ≤ 𝑅 va Gerretsen 16𝑅𝑟−5𝑟2≤𝑠2 dan: 72𝑅𝑟−9𝑟2 5≤𝑠2 20(1+𝑟/𝑅)+2⋅𝑠2−(2𝑅+𝑟)2 4𝑅2≤25+7⋅𝑠2+𝑟2−4𝑅2 4𝑅2 → 20∑𝑐𝑜𝑠𝐴+2∏𝑐𝑜𝑠𝐴≤25+7∑𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 (5 − 𝑐𝑜𝑠 𝐴) bilan ekvivalent ko‘rinish: ∑(5−𝑐𝑜𝑠𝐴)(5−𝑐𝑜𝑠𝐵)≤2 3⋅∏(5−𝑐𝑜𝑠𝐴) Shundan: ∑1 5 − 𝑐𝑜𝑠𝐴≤2 3 Mustaqil yechish uchun misollar: 1.Ixtiyoriy ∆𝐴𝐵𝐶 uchburchak uchun quyidagi tengsizlikni isbotlang: 𝑟(4𝑅+ 𝑟)≥√3𝛥 2. 𝐴𝐵𝐶 uchburchak uchun ushbu tengsizlikni isbotlang:√15 4+∑𝑐𝑜𝑠𝐴 − 𝐵≥ ∑𝑠𝑖𝑛𝐴 3.𝐴𝐵𝐶 uchburchak uchun ushbu tengsizlikni isbotlang: ∑𝑐𝑜𝑠𝐵 − 𝐶 22≥24⋅ ∏𝑠𝑖𝑛𝐴 2 4.∆𝐴𝐵𝐶 uchburchak uchun (∑𝑎𝑏)(𝑠2+𝑟2)≥4𝑎𝑏𝑐𝑠+36𝑅2𝑟2 ni isbotlang. Educational Research in Universal Sciences ISSN: 2181-3515 VOLUME 4 | SPECIAL ISSUE 12 | 2025 https://t.me/Erus_uz Multidisciplinary Scientific Journal October, 2025 135 FOYDALANILGAN ADABIYOTLAR RO‘YXATI: (REFERENCES) 1. Gerretsen, J. (1953). Inequalities between the elements of a triangle. Indagationes Mathematicae, 15, 488–499. 2. Mitrinović, D. S., Pečarić, J. E., & Volenec, V. (1989). Recent Advances in Geometric Inequalities. Dordrecht: Springer. 3. Bottema, O., Djordjević, R. Z., Janic, R. R., & Mitrinović, D. S. (1969). Geometric Inequalities. Groningen: Wolters-Noordhoff. 4. Johnson, R. A. (2007). Advanced Euclidean Geometry: Modern Methods for Classic Problems. New York: Dover Publications.