$AD: ab+a+b \geq 3\sqrt[3]{a^{2}b^{2}}, \forall a>0,b>0 $$\Rightarrow VT \geq \frac{(x+1)(Y+1)^{2}}{(z+1)(x+1)}+\frac{(y+1)(z+1)^{2}}{(x+1)(y+1)}+\frac{(z+1)(x+1)^{2}}{(y+1)(z+1)}$
$=\frac{(y+1)^{2}}{z+1}+\frac{(z+1)^{2}}{x+1}+\frac{(x+1)^{2}}{y+1}$
$\geq \frac{[(y+1)+(z+1)+(x+1)]^{2}}{(z+1)+(x+1)+(y+1)}=x+y+z+3$ (đpcm) :D