Question

Với x;y;z> 0 thoả mãn hệ thức \(x+y+z=18\sqrt{2}\)

Chứng minh rằng : \(\frac{1}{\sqrt{x\left(y+z\right)}}+\frac{1}{\sqrt{y\left(z+x\right)}}+\frac{1}{\sqrt{z\left(x+y\right)}}\ge\frac{1}{4}\)

Answer 1

Côsi: \(\sqrt{x\left(y+z\right)}=\frac{1}{2\sqrt{2}}.2.\sqrt{2x}.\sqrt{y+z}\le\frac{1}{2\sqrt{2}}\left(2x+y+z\right)\)

\(\Rightarrow\frac{1}{\sqrt{x\left(y+z\right)}}\ge\frac{2\sqrt{2}}{2x+y+z}\)

Tương tự các cái kia.

\(\Rightarrow VT\ge2\sqrt{2}\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)

\(\ge2\sqrt{2}.\frac{9}{2x+y+z+2y+z+x+2z+x+y}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)

Answer 2

\(\sum\frac{1}{\sqrt{x\left(y+z\right)}}=\sum\frac{\sqrt{2}}{\sqrt{2x}.\sqrt{y+z}}\ge\sum\frac{2\sqrt{2}}{2x+y+z}\ge2\sqrt{2}.\frac{9}{\sum\left(2x+y+z\right)}=\frac{18\sqrt{2}}{4\left(x+y+z\right)}=\frac{1}{4}\)