Question

cho a,b,c là số thực dương, tìm max: \(\sqrt{\dfrac{a}{b+c+2a}}+\sqrt{\dfrac{b}{c+a+2b}}+\sqrt{\dfrac{c}{a+b+2c}}\)

Answer 1

ad bunhiacopxki ta có

A^2 \(\le3\left(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2c}+\dfrac{c}{a+b+2c}\right)\)

Đặt B=\(\dfrac{a}{b+c+2a}+\dfrac{b}{c+a+2b}+\dfrac{c}{a+b+2c}\)

\(\Leftrightarrow\)B-3 =-\(\left(a+b+c\right)\) \(\left(\dfrac{1}{b+c+2a}+\dfrac{1}{c+a+2b}+\dfrac{1}{a+b+2a}\right)\)

dễ CM \(\dfrac{1}{a+b+2c}+\dfrac{1}{b+c+2a}+\dfrac{1}{c+a+2b}\)\(\ge\dfrac{9}{4\left(a+c+b\right)}\)

\(\Rightarrow\)B-3\(\le\)\(\dfrac{-9}{4}\)\(\Rightarrow\)B\(\le\dfrac{3}{4}\)

\(\Rightarrow A^2\le\dfrac{9}{4}\) mà A>0

\(\Rightarrow\)A\(\le\dfrac{3}{2}\)Dấu = xra khi a=b=c