Prove this inequality for given conditions For all $x,y>0$, $$\frac{1}{(x+1)^2} + \frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$$
I can only think of substituting $x+1$ with $a$ and $y+1$ with $b$. Then the inequality turns into
$$(a^2 + b^2) (ab-a-b +2) \ge a^2b^2$$
I can proceed no further. Please help.

After your substitutions we get new conditions $a>1$ and $b>1$, which makes the inequality harder. 
By the way, it's just $$xy(x-y)^2+(xy-1)^2\geq0.$$
Also, we can use C-S:
$$\sum_{cyc}\frac{1}{(x+1)^2}=\sum_{cyc}\frac{y}{(x\sqrt{y}+\sqrt{y})^2}\geq\sum_{cyc}\frac{y}{(xy+1)(x+y)}=\frac{1}{xy+1}.$$
