Question

\(\sqrt{\frac{a}{2a+b+c}}+\sqrt{\frac{b}{a+2b+c}}+\sqrt{\frac{c}{a+b+2c}}\le\frac{3}{2}\)

Answer 1

@Nguyễn Việt Lâm

Answer 2

Cần điều kiện a;b;c dương

Đặt vế trái là P, áp dụng BĐT Bunhicopxki:

\(P^2\le3\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\)

Đặt \(A=\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}=\frac{a}{a+b+a+c}+\frac{b}{a+b+b+c}+\frac{c}{a+c+b+c}\)

\(\Rightarrow A\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c}\right)=\frac{3}{4}\)

\(\Rightarrow P^2\le3.\frac{3}{4}=\frac{9}{4}\Rightarrow P\le\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)