If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? If I have that $X$ is a random variable satisfying $0\leq X \leq 1$, how can I show that $P\left(X \geq \frac{E(X)}{2}\right) \geq \frac{E(X)}{2}$? I saw a footnote which gave a hint to split up $E(X)$ as two integrals over $\left[X<\frac{E(X)}{2}\right]$ and $\left[X\geq\frac{E(X)}{2}\right]$. However, I am not quite sure how to bound this integral. Would anyone have any ideas?
 A: Write
$$
X
= X \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}
+ X \mathbf{1}_{\left\{X < \frac{E[X]}{2}\right\}},
$$
and note that
$$
X \mathbf{1}_{\left\{X < \frac{E[X]}{2}\right\}}
\leq \frac{E[X]}{2}.
$$
Also, since $0 \leq X \leq 1$, we have
$$
X \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}
\leq \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}.
$$
Then
$$
\begin{aligned}
E[X]
&= E\left[X \mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}\right]
+ E\left[X \mathbf{1}_{\left\{X < \frac{E[X]}{2}\right\}}\right] \\
&\leq E\left[\mathbf{1}_{\left\{X \geq \frac{E[X]}{2}\right\}}\right]
+ E\left[\frac{E[X]}{2}\right] \\
&= P\left(X \geq \frac{E[X]}{2}\right) + \frac{E[X]}{2}.
\end{aligned}
$$
Therefore $P\left(X \geq \frac{E[X]}{2}\right) \geq \frac{E[X]}{2}$.
A: Although the question has been already perfectly answered by Artem Mavrin, I would like to write the answer in a more measure theoretical way, specifically avoiding the introduction of new random variables ad expliciting the integrals appearing.
By definition, since $X \leq 1$,
\begin{align*}
E(X) & =\int _0^1 P(X > t) dt\\
& =\int _0^{\frac{E(X)}{2}} P(X > t) dt + \int_{\frac{E(X)}{2}}^1 P(X > t) dt.
\end{align*}
Since $P(X > t) \leq P(\Omega) = 1$ the first integral is bounded by $\frac{E(X)}{2}$, the second integral is bounded by the sup of the integrand, i.e. $P\left(X\geq \frac{E(X)}{2}\right)$.  
