$(y+\sqrt{xz}+z)^{2}\leq (x+y+z)(y+2z)$nên ta có
$\frac{2x^{2}+xy}{(y+\sqrt{xx}+z)^{2}}\geq \frac{1}{x+y+z}.(\frac{2x^{2}+xy}{y+2z})=\frac{1}{x+y+z}.(\frac{2x(x+y+z)}{y+2z}-x)$
$=\frac{2x}{y+2z}-\frac{x}{x+y+z}$
tương tự rồi cộng lại đc
$VT\geq \frac{2x}{y+2z}+\frac{2y}{z+2x}+\frac{2z}{x+2y}-1\geq \frac{2(x+y+z)^{2}}{3(xy+yz+zx)}-1\geq 1$
suy ra đpcm