Question

cho a,b,c>0 và ab+bc+ac=1. tìm min F = \(\frac{a^8}{\left(a^2+b^2\right)^2}\)+ \(\frac{b^8}{\left(b^2+c^2\right)^2}\)+ \(\frac{c^8}{\left(a^2+c^2\right)^2}\)

Answer 1

F>=a^8/2(a^4+b^4)+b^8(b^4+c^4)+c^8/(c^4+a^4)>=(a^4+b^4+c^4)^2/4(a^4+b^4+c^4)=(a^4+b^4+c^4)/4

a^2+b^2+c^2>=ab+bc+ca=1.

3(a^4+b^4+c^4)>=(a^2+b^2+c^2)^2=1>>>a^4+b^4+c^4>=1/3

>>>F>=1/3/4=1/12

Dấu = xảy ra khi a=b=c(tự tính)