Prove this inequality for given conditions

Question

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.

Answer 1

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: $$\sum_{cyc}\frac{1}{(x+1)^2}=\sum_{cyc}\frac{y}{(x\sqrt{y}+\sqrt{y})^2}\geq\sum_{cyc}\frac{y}{(xy+1)(x+y)}=\frac{1}{xy+1}.$$

Answer 2

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:

$$ xy(x^2+y^2) - x^2y^2-2xy+1 \color{blue}{\ge 2x^2y^2}-x^2y^2-2xy+1 = (xy-1)^2 \ge 0 $$