Prove that positive $x,y$ satisfy $\left(\frac{1}{1+x}\right)^2+\left(\frac{1}{1+y}\right)^2\ge\frac{1}{1+xy}$. 
Prove that positive $x,y$ satisfy $$\left(\frac{1}{1+x}\right)^2+\left(\frac{1}{1+y}\right)^2\ge\frac{1}{1+xy}$$

My teacher claims this lemma is often useful. How to prove it?
I've tried using $a^2+b^2\ge 2ab$ and $a^2+b^2\ge \frac{(a+b)^2}{2}$ and $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\ldots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\ldots+a_n)^2}{b_1+b_2+\ldots+b_n}$.

Let $x+y=2u$ and $xy=v^2$.
Hence, we need to prove that
$$\frac{2+2x+2y+x^2+y^2}{(1+x+y+xy)^2}\geq\frac{1}{1+xy}$$ or
$$(1+v^2)(2+4u+4u^2-2v^2)\geq(1+2u+v^2)^2$$ or
$$1+4u^2v^2-3v^4-2v^2\geq0$$ or
$$(1-v^2)^2+4v^2(u^2-v^2)\geq0,$$ 
which is true because $u^2\geq v^2$ it's just $(x-y)^2\geq0$.
Done!
