Distribution inequality I want to prove that if $Y\geqslant 0$ and $EY^2<\infty$, then $$\mathbf{P}(Y>0)\geqslant \frac{(EY)^2}{EY^2}.$$ I've got so far: $$DX\geqslant 0,$$ $$EY^2-(EY)^2\geqslant 0$$ We know that $\mathbf{P}(Y< 0)=0$, therefore $$EY^2-(EY)^2\geqslant \mathbf{P}(Y< 0),$$ $$EY^2-(EY)^2\geqslant EY^2\mathbf{P}(Y< 0).$$ $$EY^2-EY^2\mathbf{P}(Y< 0)\geqslant (EY)^2,$$ $$EY^2(1-\mathbf{P}(Y< 0))\geqslant (EY)^2,$$ $$EY^2\mathbf{P}(Y\geqslant 0)\geqslant (EY)^2,$$ $$\mathbf{P}(Y\geqslant 0)\geqslant \frac{(EY)^2}{EY^2}.$$ I don't know how to deal with $\mathbf{P}(Y=0)$. Advice would be appreciated.

**Hint:** Observe that $Y=Y\cdot 1_{\\{Y>0\\}}$, and apply the Cauchy-Schwarz inequality.