Spring Mathematical Tournament

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