$VT=\sum_{}^{}\sqrt{\frac{xy}{xy+z(x+y+z)}}=\sum_{}^{} \sqrt{\frac{xy}{(x+z)(y+z)}}\leq \sum_{}^{}\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{1}{2}\left(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y}\right)=\frac{3}{2}(đpcm) $Dấu $=$ xảy ra $\Leftrightarrow x=y=z=\frac{1}{3}$.
$VT=\sum_{}^{}\sqrt{\frac{xy}{xy+z(x+y+z)}}=\sum_{}^{} \sqrt{\frac{xy}{(x+z)(y+z)}}\leq \sum_{}^{}\frac{1}{2}(\frac{x}{x+z}+\frac{y}{y+z})=\frac{1}{2}(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y})=\frac{3}{2}(đpcm) $
$VT=\sum_{}^{}\sqrt{\frac{xy}{xy+z(x+y+z)}}=\sum_{}^{} \sqrt{\frac{xy}{(x+z)(y+z)}}\leq \sum_{}^{}\frac{1}{2}
\left(\frac{x}{x+z}+\frac{y}{y+z}
\right)=\frac{1}{2}
\left(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y}
\right)=\frac{3}{2}(đpcm) $
Dấu $=$ xảy ra $\Leftrightarrow x=y=z=\frac{1}{3}$.