Given $X$ is a finite non-negative discrete random variable that is not equal to identical zero. Prove that $$\Pr[X > 0] \geq \frac{E[X]^2}{E[X^2]}$$
My attempt:
$$\Pr[X > 0] = \sum_{i = 1}^n\Pr[X = i]$$
$$E[X]^2 = \left(\sum_{i = 0}^n \Pr[X = i] \cdot i\right)^2$$
$$E[X^2] = \sum_{i = 0}^n \Pr[X = i]\cdot i^2$$
$$\Pr[X > 0] \cdot E[X^2] \geq E[X]^2 \iff \sum_{i = 1}^n\Pr[X = i] \cdot \sum_{i = 0}^n \Pr[X = i]\cdot i^2 \geq \left(\sum_{i = 0}^n \Pr[X = i] \cdot i\right)^2$$
$$\left( \Pr[X = 1] + \dotso +\Pr[X = n]\right)\cdot \left( \Pr[X = 0] \cdot 0 + \Pr[X = 1] \cdot 1^2+ \dotso + \Pr[X = n] \cdot n^2\right) \geq \left( \Pr[X = 0] \cdot 0 + \Pr[X = 1] \cdot 1 + \dotso \Pr[X = n] \cdot n\right)^2$$
Also notice that $\Pr[X = i] = \Pr[X = j], \ i \neq j$, here I'm not sure about $\Pr[X = i] = \frac{1}{n}$ or $\Pr[X = i] = \frac{1}{n + 1}$, because of $0$ (intuitively it should be $\frac{1}{n + 1}$), but I think in this problem I don't need the exact value of $\Pr[X = i]$, so let $\forall i \ \Pr[X = i] = p$
$$n \cdot p (p + 4p + 9p + \dotso +n^2p ) \geq (p + 2p + \dotso + np)^2$$
What should I do next?
$X$ may not be discrete, so your argument does not work.
Note, $E(X)=E(XI(X=0))+E(XI(X>0))$
$=0+E(XI(X>0))=E(XI(X>0))\leq \sqrt{E(X^2)}\sqrt{E(I(X>0)^2)}=\sqrt{E(X^2)P(X>0)}$
Hence, $P(X>0)\geq \dfrac{(E(X))^2}{E(X^2)}$