Denote by $S$ the left-hand side of the given inequality. The arithmetic mean–geometric mean inequality implies that $xy + x + y \geq 3\sqrt[3]{x^2y^2}$ and hence
$$S \geq \frac{(x+1)(y+1)^2}{(z+1)(x+1)} + \frac{(y+1)(z+1)^2}{(x+1)(y+1)} + \frac{(z+1)(x+1)^2}{(y+1)(z+1)}.$$
Setting $a = x + 1$, $b = y + 1$, $c = z + 1$ gives
$$S \geq \frac{b^2}{c} + \frac{c^2}{a} + \frac{a^2}{b}.$$
Then
$$S \geq \frac{(a+b+c)^2}{a+b+c} = a+b+c = x+y+z+3.$$
by the Cauchy–Schwartz inequality.