tengsizlikni isbotlang

[meritocrat]

Agar a,b,c >0 bulsa quyidagi tengsizlik har doim o'rinli ekanligini isbotlang:

a^2/(a+b) + b^2/(b+c) + c^2/(c+a) >= (a+b+c)/2

[I_am_bandidos]

Hozir men darsdaman, lekin kutubhonaga borsam javobini yozaman, hozir, hozir....

[I_am_bandidos]

Ana keldim. Tak endi esa javobi:

Echim:
Avval bir tengsizlikni ko'rib chiqaylik:

(a-b)^2>=0 ni ko'rish juda oson, chunki har qanday sonning kvadrati musbatdir.

a^2+b^2>=2ab kvadratni ochib hiqib, 2ab ni o'ng tarafga olib o'tamiz.

(a+b)^2>=4ab ikki tarafga 2ab ni qo'shib chap tarafni kvadratga to'ldiramiz.

(a+b)/4>=ab/(a+b) ikki tarafni 4(a+b) ga bo'lamiz.

(*)-ab/(a+b)>=-(a+b)/4 bu operatsiyani hatto mening 10 yoshli ukam ham biladi ;). ( bu erda (*) qadamimizning ohirgi natijasining belgisi)


Keling endi berilgan tengsizlikdagi a^2/(a+b) ni tekshirib ko'raylik (qolgan 2tasi ham shu tarzda tekshirilishi mumkin):

a^2/(a+b)=(a+b)a/(a+b)-ab/(a+b)=

=a+[-ab/(a+b)]>=a-(a+b)/4= (*)dan foydalandik.

=(3a-b)/4

Demak, (**) a^2/(a+b)>=(3a-b)/4

Shunda yuqoridagi usulni qolgan ikkitasiga qo'llasak (**)giday chiqadi, shuning uchun,

a^2/(a+b) + b^2/(b+c) + c^2/(c+a) >= (3a-b)/4+(3b-c)/4+(3c-a)/4=

=(a+b+c)/2


Yahooooooo!!!! Qale, zo'rmi? Masalanichi, shunday qilib s!k!sh kere.

Bandito, are you a Maniac Mathematician? Are you planning to pursue a career in the field of Mathematics/ Actuarial etc?

[I_am_bandidos]

Dude, yeah. I am kinda math killa. I've got like 1000 math books. And the problem was not so hard. So u r kinda right, man.