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 $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\ldots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\ldots+a_n)^2}{b_1+b_2+\ldots+b_n}$.
On
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.
On
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!
Not particularly elegant, but if you multiply everything out, you get $$x^3y+xy^3+1\ge x^2y^2+2xy$$ And that's just $2$ simple AM-GM inequalities $$x^3y+xy^3\ge2x^2y^2$$ $$x^2y^2+1\ge2xy$$