<s>[INST] 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 $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\ldots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\ldots+a_n)^2}{b_1+b_2+\ldots+b_n}$.
 [/INST] One possible approach: we can divide it into cases.  
Case 1: $x,y \geq 1$ or $x,y \leq 1$
Using $a^2+b^2\ge 2ab$, which yields
$$
\left(\frac{1}{1+x}\right)^2+\left(\frac{1}{1+y}\right)^2\ge\frac{2}{1+x+y+xy}
$$
We must then show that
$$
\frac{2}{1+x+y+xy}\geq \frac{1}{1 + xy} \iff\\
2 + 2xy \geq 1 + x + y + xy \iff\\
1 - x - y + xy \geq 0 \iff\\
(x-1)(y-1) \geq 0
$$
Case 2: Otherwise
Not sure what to do here yet.  Hopefully, this partial answer can be useful.
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