TITLE: $ X>0, E(X)=1, E(X^2)=b, \forall a\in (0,1): P(X>a)\geq \frac{(1-a)^2}{b} $
QUESTION [6 upvotes]: Suppose that
$ X>0, E(X)=1, E(X^2)=b $
And We should prove for every $ a $ such that $ 0 < a < 1 $ the following statement:
$ P(X>a)\geq \frac{(1-a)^2}{b} $
This is a preliminary course of probability and we learned only basic formulas and inequalities of Markov and Chebyshev's. I would be happy if you keep the answer simple as much as possible.
Thanks.

REPLY [4 votes]: Write
$$
1=\mathsf{E}X=\mathsf{E}X1\{X>a\}+\mathsf{E}X1\{X\le a\}.
$$
Since $\mathsf{E}X1\{X\le a\}\le a$ and $\mathsf{E}X1\{X>a\}\le \sqrt{\mathsf{E}X^2\mathsf{P}(X>a)}$ by Cauchy-Schwartz,
$$
1-a\le \sqrt{\mathsf{E}X^2\mathsf{P}(X>a)},
$$
which implies the result.

REPLY [2 votes]: Let $\mathbb{1}_A$ be the indicator function of set $A$. Then for any $a\in(0,1),$ it is easy to see that
$$X \mathbb{1}_{\{X>a\}}\ge X-a.$$ Taking the expectation of both sides yields
$$E[X\mathbb{1}_{\{X>a\}}]\ge E[X]-a=1-a. \qquad(1)$$
Moreover, by the Cauchy-Schwarz inequality,
$$E[X \mathbb{1}_{\{X>a\}}]\le \sqrt{E[X^2]}\sqrt{P\{X>a\}}. \qquad(2)$$
Finally, from $(1)$ and $(2)$, we deduce
$$\sqrt{E[X^2]}\sqrt{P\{X>a\}}\ge 1-a \implies P\{X>a\}\ge \frac{(1-a)^2}{E[X^2]}=\frac{(1-a)^2}{b}.$$