$P[X\geq m/2]\geq\frac{1}{m+1}$ if $\mathbb{E}[X]=\frac{m}{2}$ Let $m\gt 0$ and let $X$ be a random variable on the set $\{0,\dots,m\}$.
Assuming that $\mathbb{E}[X]=\frac{m}{2}$, prove that:
$$P[X\geq m/2]\geq\frac{1}{m+1}$$
I tried induction but it didn't work, know I am trying to prove it by contradiction, someone got a hint?

If $P(X \ge m/2) < \frac{1}{m+1}$, then
\begin{align}
E[X] &= \sum_{k < m/2} k P(X=k) + \sum_{k \ge m/2} kP(X=k)
\\
&\le \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) P(X < m/2) + m P(X \ge m/2)
\\
&= \left(\left\lceil\frac{m}{2}\right\rceil - 1\right)(1-P(X \ge m/2)) + mP(X \ge m/2)
\\
&= \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) +
\left(m+1 - \left\lceil\frac{m}{2}\right\rceil\right) P(X \ge m/2)
\\
&{\color{blue}<} \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) +
\left(m+1 - \left\lceil\frac{m}{2}\right\rceil\right) \frac{1}{m+1}
\\
&= \left\lceil\frac{m}{2}\right\rceil \frac{m}{m+1}
\\
&= \begin{cases}
\frac{m}{2} \cdot \frac{m}{m+1} & \text{$m$ even}
\\
\frac{m+1}{2} \cdot \frac{m}{m+1} & \text{$m$ odd}
\\
\end{cases}
\\
&\le \frac{m}{2},
\end{align}
a contradiction.
Note that the statement is true even if $E[X]=m/2$ is replaced with $E[X] \ge m/2$.
Intuition: we want to see how small $P(X \ge m/2)$ can be while still keeping the expectation $E[X]$ large. Thus we should consider random variables that only take values $\lceil m/2 \rceil - 1$ and $m$. This is the first inequality in the work above.
