Lower bound for $Pr(X > E[X])$ where $X$ is a non-negative integer random variable

We have a random variable X which is a non-negative integer with bounds $0 \leq X \leq m$ where $m$ is also an integer and a multiple of 3.

We also know the expected value of $X$: $E[X]$ = $\frac{2m}{3}$.

Is it possible to prove that $Pr[X \geq E[X]] \geq \frac{1}{m}$?

I tried using Markov's inequality but only found a lower bound for $Pr[X > \frac{m}{3}] \geq \frac {1}{2}$.

$$\begin{align*} \mathbb{E}[X]&=\mathbb{E}[X1_{\{X\geq\mathbb{E}[X]\}}]+\mathbb{E}[X1_{\{X<\mathbb{E}[X]\}}]\\ &\leq m\mathbb{P}(X\geq\mathbb{E}[X])+\mathbb{E}[X]-1\\ &\Rightarrow\mathbb{P}(X\geq\mathbb{E}[X])\geq\frac{1}{m}. \end{align*}$$ Here the inequality uses $X\leq m$ and the fact that $m$ is a multiple of 3 hence $\mathbb{E}[X]$ is integer, together with $X\in\mathbb{Z}$ implies $X1_{\{X<\mathbb{E}[X]\}}\leq\mathbb{E}[X]-1$.