Estimating for non negative r.v. $P[X>0]$

Question

I am currently independently studying elementary measure theory, and I recently came across a proof that made use of the fact that for $X$ a non-negative random variable, $$P[X>0]>\frac{E[X]^2}{4E[X^2]}$$ Me being a skeptic, I tried to prove this. The only thing I have left to show is $$P[X>0]>E[X]-E[X^2]$$ as if you believe this, then for $a\in \mathbb{R}$, $P[aX>0]=P[X>0]\implies P[X>0]>aE[X]-a^2E[X^2]$ (by linearity of expectation). Now simply use a bit of calc, and find $a= E[X]/2E[X^2]$. Substitute in and you get the desired lower bound. Any help is greatly appreciated.

Answer 1

There's actually a tighter bound. $$P(X>0) \geq \frac{E[X]^2}{E[X^2]} \geq \frac{E[X]^2}{4E[X^2]}$$

To see this, note that for any non-negative random variable:

$$X = X1_{X>0} \Rightarrow E[X] =E[X1_{X>0}]\leq \sqrt{E[X^2]P[X>0]} $$

Last step is by Cauchy Schwarz inequality. $1_{(.)}$ is the indicator function.