If $X\ge 0$ and $a\ge E[X]$, then $P(X\gt a)\ge (E(X)-a)^2/ E(X^2)$

Question

I need help with this problem.

Prove that if $X\ge0$ and $E[X^2]<\infty$ then for all $a\neq0$, $E[X] > a$, we have $$P(X\gt a)\ge\frac {(E(X)-a)^2}{E(X^2)}$$

Progress

I have my doubts if this problem is true, but I need your help to confirm it.

Edit:

I think the problem works when we set $E[X] > a $. Previously it was $E[X] \leq a $.

Answer 1

As stated, the statement seems false. Tacking $X$ to be a Bernoulli with parameter $p=1$ ($X=1$ a.s.; $\mathbb{E}[X]=\mathbb{E}[X^2] = 1$), that would mean that for all $a\geq 1$ $$ 0 = \Pr[X > a] \geq (a-1)^2 \geq 0\ . $$