Question

Cho a,b,c>0 thỏa mãn a+b+c=1 CMR \(\dfrac{a}{\sqrt{a+bc}}+\dfrac{b}{\sqrt{b+ca}}+\dfrac{c}{\sqrt{c+ab}}\le\dfrac{3}{2}\)

Answer 1

\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\Rightarrow\dfrac{a}{\sqrt{a+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{\dfrac{a}{a+b}+\dfrac{a}{a+c}}{2}\)

\(\Rightarrow\Sigma\dfrac{a}{\sqrt{a+bc}}=\Sigma\dfrac{a}{\sqrt{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{a}{a+c}+\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{c+a}+\dfrac{c}{c+b}}{2}=\dfrac{3}{2}\)

\(dấu"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)