Câu hỏi của trung quoc

Question

cho a>=0;b>=0, chung minh a+b+$\frac{1}{2}$>=$\sqrt{a}+\sqrt{b}$

Nguyễn Linh Chi · Accepted Answer

Ta có:

$a+b+\frac{1}{2}=\left(a+\frac{1}{4}\right)+\left(b+\frac{1}{4}\right)\ge2\sqrt{a.\frac{1}{4}}+2.\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}$

Dấu "=" xảy ra <=> a = b = 1/4

Câu trả lời:

Ta có:

$a+b+\frac{1}{2}=\left(a+\frac{1}{4}\right)+\left(b+\frac{1}{4}\right)\ge2\sqrt{a.\frac{1}{4}}+2.\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}$

Dấu "=" xảy ra <=> a = b = 1/4