Team Selection Test

Note that $(x+y{)}^2\le 2(x^2+y^2)$ and $4z+4\le 2(z^2+3)$. Therefore, $$\frac{x^2+1}{(x+y{)}^2+4(z+1)}\ge \frac{x^2+1}{2(x^2+y^2+z^2+3)}$$ Similarly we can write analogous inequalities for pairs $(y,z)$ and $(z,x)$. The sum of these three inequalities yields $$\begin{gathered} \frac{x^2+1}{(x+y{)}^2+4(z+1)}+\frac{y^2+1}{(y+z{)}^2+4(x+1)}+\frac{z^2+1}{(z+x{)}^2+4(y+1)} \\ \ge \frac{x^2+y^2+z^2+3}{2(x^2+y^2+z^2+3)}=\frac{1}{2}. \end{gathered}$$

The equality holds at $x=y=z=1$.