Ta có: $\sum_{cyc}^{}\frac{x}{x+\sqrt{(x+y)(z+x)}}\leq \sum_{cyc}^{}\frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\sum\sqrt{x}=1 $
Ta có: $\sum_{cyc}^{}\frac{x}{x+\sqrt{(x+y)(z+x)}}\leq \sum_{cyc}^{}\frac{x}{x+\sqrt{xz}+\sqrt{xy}}\leq\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\sum\sqrt{x}=1 $
Ta có: $\sum_{cyc}^{}\frac{x}{x+\sqrt{(x+y)(z+x)}}\leq \sum_{cyc}^{}\frac{x}{x+\sqrt{xz}+\sqrt{xy}}
=\frac{1}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\sum\sqrt{x}=1 $