TRĐZ hêTa Có
$VT=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+3y}}+\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y+3x}}$
Ta có
$ \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+3y}}=\frac{\sqrt{x}.\sqrt{x+y}}{\sqrt{x+y}.\sqrt{x+3y}}+\frac{\sqrt{2y}}{\sqrt{x+3y}.\sqrt{2}}\leq \frac{1}{2}\left ( \frac{x}{x+y}+\frac{x+y}{x+3y} \right )+\frac{1}{2}\left ( \frac{2y}{x+3y} +\frac{1}{2}\right )=\frac{1}{2}\left ( \frac{3}{2}+\frac{x}{x+y} \right )$
với chiêu thức tương tự ta chứng minh được
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{y+3x}}\leq \frac{1}{2}\left ( \frac{y}{x+y}+\frac{3}{2} \right )$
$\Rightarrow VT\leq 2$
mọi dấu $=$ xẩy ra $x=y$