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 eliminating denominators and simplifying, the inequality reduces to:
$$
x^3y+xy^3-x^2y^2-2xy+1 \ge 0 \;\;\iff\;\; xy(x^2+y^2) - x^2y^2-2xy+1 \ge 0
$$
Using that $\,xy \ge 0\,$ and $\,\color{blue}{x^2+y^2 \ge 2xy}\,$, the above follows from:
$$
xy(x^2+y^2) - x^2y^2-2xy+1 \color{blue}{\ge 2x^2y^2}-x^2y^2-2xy+1 = (xy-1)^2 \ge 0
$$
