prove that $‎‎P(X=0) \leq ‎\frac{Var(X)}{‎‎‎E(X‎^{2})‎}‎$. Let $X$ be a random variable taking integral nonnegative values, let $E [X^2]$
denote the expectation of its square, and let $Var [X]$ denote its variance. Prove
that $‎‎P(X=0) \leq ‎\frac{Var(X)}{‎‎‎E(X‎^{2})‎}‎$‎.
I try to use this theorem which will be proved by Chebyshev's Inequality:$‎‎P(X=0) \leq ‎\frac{Var(X)}{‎‎‎E(X‎)^{2}‎}$, I try to prove this $E(X‎^{2}) \geq E(X‎)^{2} $ and using above inequality it will be done. but I couldn't prove this,I thought that using the definition of expectation will help but I am not sure,please help me with some guidance or hint or any reference,thank you very much.

Since $\text{Var}[X] \geq 0$ (this is a well-known fact), $E[X^2] - (E[X])^2 \geq 0$ and the inequality desired follows.
