Understanding proof that $P(X=0) \leq \frac{Var(X)}{E(X^2)}$

Question

I want to understand how to show that for a random variable X which takes non-negative values, it holds that $$P(X=0) \leq \frac{\textrm{Var}(X)}{E(X^2)}$$

There already is a solution to that question, but I can't see why the inequality immediately follows when I know that $\textrm{Var}(X) \geq 0$

prove that $‎‎P(X=0) \leq ‎\frac{Var(X)}{‎‎‎E(X‎^{2})‎}‎$.

Answer 1

Remark that $\mathbb{E}[X] = \mathbb{E}[X1_{X>0}]$ and apply Cauchy-Schwarz inequality: $$\mathbb{E}[X]^2\leq \mathbb{E}[X^2]\mathbb{E}[1_{X>0}] = \mathbb{E}[X^2](1-\mathbb{P}(X=0)). $$ Then conclude $$\mathbb{P}(X=0) \leq 1 - \frac{\mathbb{E}[X]^2}{\mathbb{E}[X^2]} = \frac{\text{Var} X}{\mathbb{E}[X^2]}. $$