Application of AM-GM inequality to specific contest problem Suppose that $x,y\in [0,1]$. Prove that $\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\leq \frac{2}{\sqrt{1+xy}}.$
I suppose that this problem can be solved by some application of AM-GM inequality. I was trying to do the following: since $xy\leq \frac{x^2+y^2}{2}$ then $\frac{2}{\sqrt{1+xy}}\geq \frac{2}{\sqrt{1+x^2/2+y^2/2}}$. But the inequality $\frac{2}{\sqrt{1+x^2/2+y^2/2}}\geq \frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}$ is obviously false. So I guess we have to use something which is non-trivial.
Would be grateful if someone can show the solution.
I have spent probably 2-3 hours and did not get it.

We have
$$\left(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\right)^2 = \frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{2}{\sqrt{(1+x^2)(1+y^2)}}$$
Using the AM-GM we have
$$\frac{2}{\sqrt{(1+x^2)(1+y^2)}} \leqslant \frac{1}{1+x^2}+\frac{1}{1+y^2}.$$
Therefore, we need to prove
$$\frac{1}{1+x^2}+\frac{1}{1+y^2}\leqslant \frac{2}{1+xy},$$
equivalent to
$$\frac{(xy-1)(x-y)^2}{(1+x^2)(1+y^2+1)(1+xy)} \leqslant 0.$$
Which is true.
