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.

From Cauchy-Schwarz inequality:
$\Bbb{E}(X)=\Bbb{E}(1_{\{X>0\}}X)\leq \sqrt{\Bbb{E}(1_{\{X>0\}}^2)}\cdot \sqrt{\Bbb{E}(X^2)} = \sqrt{P(X>0)}\cdot \sqrt{\Bbb{E}(X^2)}$. Therefore
$\Bbb{E}(X)^2 \leq P(X>0)\cdot \Bbb{E}(X^2)$
