Yilda matematika, Nesbittniki tengsizlik ijobiy haqiqiy sonlar uchun a, b va v,
![frac {a} {b + c} + frac {b} {a + c} + frac {c} {a + b} geq frac {3} {2}.](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec8cddcaf162a1509f22053e879df555a0c28ab1)
Bu qiyin va juda o'rganilgan elementar maxsus holat (N = 3) Shapiro tengsizligi, va kamida 50 yil oldin nashr etilgan.
Tegishli yuqori chegara yo'q, chunki tengsizlikning har qanday 3 fraktsiyasi o'zboshimchalik bilan katta bo'lishi mumkin.
Isbot
Birinchi dalil: AM-HM tengsizligi
Tomonidan AM -HM tengsizlik
,
![frac {(a + b) + (a + c) + (b + c)} {3} geq frac {3} { displaystyle frac {1} {a + b} + frac {1} {a + c} + frac {1} {b + c}}.](https://wikimedia.org/api/rest_v1/media/math/render/svg/d50d2381318c98aae99c6d04c997ab723a084706)
Nominallarni tozalash hosil
![((a + b) + (a + c) + (b + c)) left ( frac {1} {a + b} + frac {1} {a + c} + frac {1} { b + c} right) geq 9,](https://wikimedia.org/api/rest_v1/media/math/render/svg/d78ca4c4d8af7c84bfa7fe05ad46af03ec1c3fe2)
biz undan olamiz
![2 frac {a + b + c} {b + c} +2 frac {a + b + c} {a + c} +2 frac {a + b + c} {a + b} geq9](https://wikimedia.org/api/rest_v1/media/math/render/svg/25028d1ec4085ff8b7689da1d8cf8d3bc220bbf5)
mahsulotni kengaytirish va maxrajga o'xshash yig'ish orqali. Bu to'g'ridan-to'g'ri yakuniy natijaga qadar soddalashtiradi.
Ikkinchi dalil: Qayta tartibga solish
Aytaylik
, bizda shunday
![frac 1 {b + c} ge frac 1 {a + c} ge frac 1 {a + b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dab7606fbd34e0ddf57a20d8b5784cf711e4a241)
aniqlang
![vec x = (a, b, c)](https://wikimedia.org/api/rest_v1/media/math/render/svg/74c8cf41f8da636d22ee3e5442cd620279c63e80)
![vec y = chap ( frac 1 {b + c}, frac 1 {a + c}, frac 1 {a + b} o'ng)](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bcb0b6f70a864d09d74ba570dbbb852ea2c2153)
Ikki ketma-ketlikning skaler ko'paytmasi, chunki qayta tashkil etish tengsizligi agar ular xuddi shu tarzda joylashtirilgan bo'lsa, qo'ng'iroq qiling
va
vektor
bitta va ikkitadan siljigan bizda:
![vec x cdot vec y ge vec x cdot vec y_1](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bc9bdb34f232abd401e43fae13bf8d97e1a76ca)
![vec x cdot vec y ge vec x cdot vec y_2](https://wikimedia.org/api/rest_v1/media/math/render/svg/76050040cc8890cadfb01bdfa96b6e4524e07948)
Qo'shish biz istagan Nesbitt tengsizligini keltirib chiqaradi.
Uchinchi dalil: Kvadratchalar yig'indisi
Quyidagi o'ziga xoslik hamma uchun to'g'ri keladi ![a, b, c:](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f2cfbf58f4d0e093b4b94fbc4c5cdd5d88b5301)
![frac {a} {b + c} + frac {b} {a + c} + frac {c} {a + b} = frac {3} {2} + frac {1} {2} chap ( frac {(ab) ^ 2} {(a + c) (b + c)} + frac {(ac) ^ 2} {(a + b) (b + c)} + frac { (bc) ^ 2} {(a + b) (a + c)} o'ng)](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb25b43cda866b24d1cf64148a109ba93a883159)
Bu chap tomonning kam emasligini aniq isbotlaydi
ijobiy a, b va c uchun.
Izoh: har qanday ratsional tengsizlikni uni tegishli kvadratchalar identifikatoriga o'zgartirish orqali ko'rsatish mumkin, qarang Hilbertning o'n ettinchi muammosi.
To'rtinchi dalil: Koshi-Shvarts
Ga qo'ng'iroq qilish Koshi-Shvarts tengsizligi vektorlarda
hosil
![((b + c) + (a + c) + (a + b)) left ( frac {1} {b + c} + frac {1} {a + c} + frac {1} { a + b} right) geq 9,](https://wikimedia.org/api/rest_v1/media/math/render/svg/180f0e25a4123633e20f64b95b0e92e88ab01c91)
bu biz qilganimizdek yakuniy natijaga aylantirilishi mumkin AM-HM dalil.
Beshinchi dalil: AM-GM
Ruxsat bering
. Keyin biz amal qilamiz AM-GM tengsizligi quyidagilarni olish uchun
![frac {x + z} {y} + frac {y + z} {x} + frac {x + y} {z} geq6.](https://wikimedia.org/api/rest_v1/media/math/render/svg/550528b93e8a401f963c612ec81b13ae32123f6b)
chunki ![{ displaystyle { frac {x} {y}} + { frac {z} {y}} + { frac {y} {x}} + { frac {z} {x}} + { frac {x} {z}} + { frac {y} {z}} geq 6 { sqrt [{6}] {{ frac {x} {y}} cdot { frac {z} {y }} cdot { frac {y} {x}} cdot { frac {z} {x}} cdot { frac {x} {z}} cdot { frac {y} {z}} }} = 6.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1507dc4a3297ab152b5f0dcf02fe8aa41ef56045)
O'rniga
foydasiga
hosil
![{ displaystyle { frac {2a + b + c} {b + c}} + { frac {a + b + 2c} {a + b}} + { frac {a + 2b + c} {c + a}} geq 6}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a35b8cedddc6de35017b415eb11427e15a7a0b2)
![{ displaystyle { frac {2a} {b + c}} + { frac {2c} {a + b}} + { frac {2b} {a + c}} + 3 geq 6}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3e8b4157752b72b3532bcb1bf89572bc2de7d6a)
bu yakuniy natijaga qadar soddalashtiradi.
Oltinchi dalil: Titu lemmasi
Titu lemmasi, ning to'g'ridan-to'g'ri natijasi Koshi-Shvarts tengsizligi, har qanday ketma-ketligi uchun
haqiqiy raqamlar
va har qanday ketma-ketligi
ijobiy raqamlar
,
. Biz uning uch muddatli instansiyasidan foydalanamiz
-natija
va
-natija
:
![frac {a ^ 2} {a (b + c)} + frac {b ^ 2} {b (c + a)} + + frac {c ^ 2} {c (a + b)} geq frac {(a + b + c) ^ 2} {a (b + c) + b (c + a) + c (a + b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9761fbf5643111567537904b6babf2e8d24685f6)
Barcha mahsulotlarni kamroq tomonga ko'paytirib va shunga o'xshash atamalarni yig'ib, biz olamiz
![frac {a ^ 2} {a (b + c)} + frac {b ^ 2} {b (c + a)} + + frac {c ^ 2} {c (a + b)} geq frac {a ^ 2 + b ^ 2 + c ^ 2 + 2 (ab + bc + ca)} {2 (ab + bc + ca)},](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a4066e5bee3eb4f3482c0d81958dc366e35b86d)
bu soddalashtiradi
![frac {a} {b + c} + frac {b} {c + a} + frac {c} {a + b} geq frac {a ^ 2 + b ^ 2 + c ^ 2} { 2 (ab + bc + ca)} + 1.](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d15148e20ca000f7539792a64f70b5e795842f2)
Tomonidan qayta tashkil etish tengsizligi, bizda ... bor
, shuning uchun kichik tomonidagi kasr kamida bo'lishi kerak
. Shunday qilib,
![frac {a} {b + c} + frac {b} {c + a} + frac {c} {a + b} geq frac {3} {2}.](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ad3400e459831ddaaa20fd8291a121637a2270)
Ettinchi dalil: Bir hil
Tengsizlikning chap tomoni bir hil bo'lgani uchun, biz taxmin qilishimiz mumkin
. Endi aniqlang
,
va
. Istalgan tengsizlik aylanadi
, yoki teng ravishda,
. Bu Tituning "Lemma" sida aniq.
Sakkizinchi dalil: Jensen tengsizligi
Aniqlang
va funktsiyasini ko'rib chiqing
. Ushbu funktsiyani konveks sifatida ko'rsatish mumkin
va chaqirish Jensen tengsizligi, biz olamiz
![{ displaystyle displaystyle { frac {{ frac {a} {Sa}} + { frac {b} {Sb}} + { frac {c} {Sc}}} {3}} geq { frac {S / 3} {SS / 3}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f0fcf805cc7e5879ff3b9ae3ed41902ac615597)
To'g'ridan to'g'ri hisoblash hosil beradi
![frac {a} {b + c} + frac {b} {c + a} + frac {c} {a + b} geq frac {3} {2}.](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ad3400e459831ddaaa20fd8291a121637a2270)
To'qqizinchi dalil: Ikki o'zgaruvchili tengsizlikni kamaytirish
Nomzodlarni tozalash orqali,
![{ displaystyle { frac {a} {b + c}} + { frac {b} {a + c}} + { frac {c} {a + b}} geq { frac {3} { 2}} iff 2 (a ^ {3} + b ^ {3} + c ^ {3}) geq ab ^ {2} + a ^ {2} b + ac ^ {2} + a ^ {2 } c + bc ^ {2} + b ^ {2} c.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fca19656a680d8c9952caea533c8ddf94423ff33)
Endi buni isbotlash kifoya
uchun
, buni uch marta yig'ish uchun
va
dalilni to'ldiradi.
Sifatida
biz tugatdik.
Adabiyotlar
- Nesbitt, AM, Muammo 15114, Education Times, 55, 1902.
- Ion Ionesku, Ruminiya matematik gazetasi, XXXII jild (1926 yil 15 sentyabr - 1927 yil 15 avgust), 120-bet
- Artur Lohuoter (1982). "Tengsizliklarga kirish". PDF formatidagi onlayn elektron kitob.
Tashqi havolalar