Consider a random variable $X$ taking values in the positive integers. If
\begin{equation} \tag 1 P(X ≤ 2E[X]) = 1, \end{equation} Can we prove that $P\left(X \leq \frac{E[X]}{2}\right) \leq \frac{2}{3}$?
From (1) and Markov's Inequality $\left(P(X \geq a)\leq \frac{E[X]}{a}\right)$ for $a=2E[X]$ we can see $P\left(X=2E[X]\right)\leq \frac{1}{2}$. We can use Markov's Inequality for $a=\frac{E[X]}{2}$, but we get to an obvious statement. Here is where I stuck. I don't see any more information to use and go further.
You don't require $\ X\ $ to take values in the positive integers, but merely that $\ E(X)\ $ be finite and positive. Let $\ \alpha=$$P\left(X\le\frac{E(X)}{2}\right)\ $. Then $\ P\left(\frac{E(X)}{2}< X\le2E(X)\right)=$$1-\alpha\ $, and \begin{align} E(X)&=\int_{\left(-\infty,2E(X)\right]}xdP(x)\\ &= \int_{\left(-\infty,\frac{E(X)}{2}\right]}xdP(x)+ \int_{\left(\frac{E(X)}{2},2E(X)\right]}xdP(x)\\ &\le \frac{\alpha E(X)}{2}+2(1-\alpha)E(X)\ , \end{align} from which, if $\ E(X)>0\ $, it follows that $\ \alpha\le\frac{2}{3}\ $.