Prove that $E\left(\frac{XY}{X^2+Y^2}\right) \geqslant 0$ for i.i.d. $X$ and $Y$ 
Let $X$ and $Y$ be two independent identically distributed random
  variables. Prove that
  $$E\left(\frac{XY}{X^2+Y^2}\right) \geqslant 0.$$

I tried to manipulate with the expression $\dfrac{XY}{X^2+Y^2}$, e.g. tried to rewrite it in the form 
$$\dfrac{XY}{X^2+Y^2} = \dfrac{(X+Y)^2}{4(X^2+Y^2)} - \dfrac{(X-Y)^2}{4(X^2+Y^2)}$$
or even in the form:
$$\dfrac{XY}{X^2+Y^2} = \dfrac{(X+Y)^2 - (X-Y)^2}{2(X+Y)^2 + 2(X-Y)^2}$$
but it didn't help much.
I am interested to see a proof of this fact or some favorable ideas that may help here.
Reference: The problem was proposed in the Kolmogorov Students' Competition in Probability Theory, 2017.
 A: Here is a proof:
$$ \Bbb{E}\left[\frac{XY}{X^2+Y^2}\right]
= \Bbb{E}\left[ \int_{0}^{\infty} XY e^{-(X^2+Y^2)t} \, dt \right]
= \int_{0}^{\infty} \Bbb{E}[X e^{-X^2 t}]\Bbb{E}[Y e^{-Y^2 t}] \, dt \geq 0. $$

Addendum. My observation is that the expectation can be thought as a quadratic form on the space of finite measures. Indeed, if $\mu$ is the common law of $X$ and $Y$, then
$$ \Bbb{E}\left[ \frac{XY}{X^2+Y^2}\right] = \int_{\Bbb{R}^2} \frac{xy}{x^2+y^2} \, \mu(dx)\mu(dy). $$
So the problem boils down to showing that this form is positive semi-definite. One such way is to find a unitary transformation under which the quadratic form reduces to the usual $2$-norm. Such a transform are often given as integral transformation:
$$ L\mu(t) = \int_{\Bbb{R}} k(t, x) \, \mu(dx). $$
Assuming that $L$ provides such a transformation, it should satisfy
$$ \int_{\Bbb{R}^2} \frac{xy}{x^2+y^2} \, \mu(dx)\mu(dy)
= \int_{\Bbb{R}} (L\mu(t))^2 \, dt
= \int_{\Bbb{R}^2} \left(\int_{\Bbb{R}} k(t, x)k(t, y) \, dt \right) \mu(dx)\mu(dy). $$
When the dust settles down, it becomes clear what we should hunt: a function $k$ such that
$$ \frac{xy}{x^2+y^2} = \int_{\Bbb{R}} k(t, x)k(t, y) \, dt. $$
It is not hard to guess such a function, and indeed my proof uses $k(t, x) = x e^{-x^2 t} \mathbf{1}_{\{t \geq 0\}}$.
