$2(x^2+y^2) \geq (x+y)^2$ and $(z-1)^2\geq 0 \rightarrow 2z^2+6\geq 4(z+1)$, these inequalities give: $$\rightarrow \frac {x^2+1}{(x+y)^2+4(z+1)} \geq \frac {x^2+1}{2x^2+2y^2+2z^2+6}$$$$\rightarrow \frac {y^2+1}{(y+z)^2+4(x+1)} \geq \frac {y^2+1}{2x^2+2y^2+2z^2+6}$$$$\rightarrow \frac {z^2+1}{(z+x)^2+4(y+1)} \geq \frac {z^2+1}{2x^2+2y^2+2z^2+6}$$$$\rightarrow \frac {x^2+1}{(x+y)^2+4(z+1)} + \frac {y^2+1}{(y+z)^2+4(x+1)} + \frac {z^2+1}{(z+x)^2+4(y+1)} \geq \frac{x^2+y^2+z^2+3}{2x^2+2y^2+2z^2+6} = \frac {1}{2}$$

$[(x+y)^2+4(z+1)]+[(y+z)^2+4(x+1)]+[(z+x)^2+4(y+1)]=(x+y+1)^2+(y+z+1)^2+(z+x+1)^2+9$ Let's say : $A=(x+y+1)^2+(y+z+1)^2+(z+x+1)^2$ $B=\frac {x^2+1}{(x+y)^2+4(z+1)} + \frac {y^2+1}{(y+z)^2+4(x+1)} + \frac {z^2+1}{(z+x)^2+4(y+1)}$ By Titu's Lemma: $$\frac {x^2}{(x+y)^2+4(z+1)} + \frac{y^2}{(y+z)^2+4(x+1)} + \frac{1}{(z+x)^2+4(y+1)} \geq \frac {(x+y+1)^2}{A+9}$$$$\frac {y^2}{(y+z)^2+4(x+1)} + \frac{z^2}{(z+x)^2+4(y+1)} + \frac{1}{(x+y)^2+4(z+1)} \geq \frac {(y+z+1)^2}{A+9}$$$$\frac {z^2}{(z+x)^2+4(y+1)} + \frac{x^2}{(x+y)^2+4(z+1)} + \frac{1}{(y+z)^2+4(x+1)} \geq \frac {(z+x+1)^2}{A+9}$$$$\frac {1}{(x+y)^2+4(z+1)} + \frac{1}{(y+z)^2+4(x+1)} + \frac{1}{(z+x)^2+4(y+1)} \geq \frac {9}{A+9}$$$$\implies 2B \geq \frac{A+9}{A+9}=1 \implies B\geq \frac{1}{2}$$$$\implies \frac {x^2+1}{(x+y)^2+4(z+1)} + \frac {y^2+1}{(y+z)^2+4(x+1)} + \frac {z^2+1}{(z+x)^2+4(y+1)} \geq \frac{1}{2}$$