Question

Cho \(a,b,c\ge0\)

Tìm \(P_{min}=\Sigma_{cyc}\left(\sqrt[4]{\frac{a}{b+c}}+\frac{b+c}{a}\right)\)

 

Answer 1

nguyem,ưdjxrckrk

Answer 2

Áp dụng cosi ta có 

\(\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{a}{b+c}}+\sqrt[4]{\frac{a}{b+c}}+\frac{b+c}{2a\sqrt[4]{2}}\ge5\sqrt[5]{\frac{1}{\sqrt[4]{2^5}}}=\frac{5}{\sqrt[4]{2}}\)

Khi đó

\(4P\ge\frac{15}{\sqrt[4]{2}}+\left(4-\frac{1}{2\sqrt[4]{2}}\right)\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\)

Mà \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=\left(\frac{a}{b}+\frac{b}{a}\right)+...\ge6\)

=> \(4P\ge\frac{15}{\sqrt[4]{2}}+\left(4-\frac{1}{2\sqrt[4]{2}}\right).6=24+\frac{12}{\sqrt[4]{2}}\)

=> \(P\ge6+\frac{3}{\sqrt[4]{2}}\)

dấu bằng xảy ra khi a=b=c