Show that $P(Y> 0)\ge E[Y]^2/{E}[Y^2]$

Question

Let $Y$ be an integrable random variable such that $E(Y)>0$. Then $$P(Y> 0)\ge E[Y]^2/{E}[Y^2]$$

I am trying to figure this inequality out. My idea is to find a polynomial $p(Y)$ of second order such that $$1_{Y>0} \ge p(Y)$$ then to take the expectation in both sides giving $$P(Y> 0)\ge E[p(Y)]$$ and the polynomial must be like $E[p(Y)] = E[Y]^2/{E}[Y^2]$ or at least $E[p(Y)] \ge E[Y]^2/{E}[Y^2]$

Answer 1

Hint: If $Y$ has finite variance, by cauchy-schwarz-inequality,

$$ \operatorname{E}[Y] \le \operatorname{E}[ Y \, \mathbf{1}_{\{ Y > 0 \}} ] \le \operatorname{E}[Y^2]^{1/2} \operatorname{P}( Y > 0)^{1/2}. $$

If $E[Y^2] = \infty$, since $E[Y] < \infty$, then the inequality holds trivially.