How can I prove that $\Bbb{P}(X>0)\geq \frac{\Bbb{E}(X)^2}{\Bbb{E}(X^2)}$ I have the following problem:

Let $X$ be a nonnegative random variable such that $\Bbb{E}(X^2)<\infty$ Then show that $$\Bbb{P}(X>0)\geq \frac{\Bbb{E}(X)^2}{\Bbb{E}(X^2)}$$

I would like to get some hints, because I did the following:
Let me define $U:=X$ and $V:=\Bbb{1}_{\{X>0\}}$. Then clearly $U\in L^2$. But also $V\in L^2$ since $$\int_\Omega (\Bbb{1}_{\{X>0\}})^2 \Bbb{P}(d\omega)=\int_\Omega \Bbb{1}_{\{X>0\}}\Bbb{P}(d\omega)=\Bbb{P}(X>0)\leq 1$$In addition we have that $\Bbb{E}(V^2)=\Bbb{E}((\Bbb{1}_{\{X>0\}})^2)=\Bbb{E}(\Bbb{1}_{\{X>0\}})=\Bbb{P}(X>0)$. Now applying the Cauchy-Schwarz inequality we get $\Bbb{E}(|UV|)^2\leq \Bbb{E}(U^2)\cdot\Bbb{E}(V^2)=\Bbb{E}(U^2)\cdot\Bbb{P}(X>0)$ so $$\Bbb{P}(X>0)\geq \frac{\Bbb{E}(|\Bbb{1}_{\{X>0\}}X|)^2}{\Bbb{E}(X^2)}$$
Now I see that $\Bbb{E}(|\Bbb{1}_{\{X>0\}}X|)=\Bbb{E}(\Bbb{1}_{\{X>0\}}X)=\Bbb{E}(X)$ since the lebesgue mesure doesn't see the nullsets.
Is this correct like this?
Thanks for your help.

Yes. Just to mention that it would also be true if we only had $\mathbb E[X]\ge0$. Indeed, since $\mathbb E[X]\le\mathbb E[X1_{\{X>0\}}]$, you could just square the inequality and prove the claim.
