已知a,b∈R+,求证:1/2(a+b)^2+1/4(a+b)≥a根号b+b根号a
问题描述:
已知a,b∈R+,求证:1/2(a+b)^2+1/4(a+b)≥a根号b+b根号a
答
啊
答
(a+b)^2/2+(a+b)/4
=(a^2+b^2)/2+ab+(a+b)/4
>=2ab+(a+b)/4
=(a+4ab+b+4ab)/4
>=(2根号(4a^2b)+2根号(4ab^2))/4
=(4a根号b+4b根号a)/4
=a根号b+b根号2
所以(a+b)^2/2+(a+b)/4>=a根号b+b根号a