How to compute a non-trivial lower bound on $P \left[|X| > \frac{|\alpha|}{2} \right]$? Let $X$ be a random variable such that $E[X] = \alpha$,  $\alpha \in \mathbb{R}$ and $E[X^2] = \beta$. The  problem is to find a lower bound on the  following probability
$$
P \left[|X| > \frac{|\alpha|}{2} \right].
$$
Any help would be appreciated.

You are going to minimise the probability in question if you put as much of the probability as possible at $\frac \alpha 2$ and the rest at a single other point, though you may need two other points when $\alpha=0$.
This happens when  $X=\frac{\alpha}{2}$ with probability $\frac{4\beta-4\alpha^2}{4\beta-3\alpha^2}$ and $X=\frac{2\beta-\alpha^2}{\alpha}$ with probability $\frac{\alpha^2}{4\beta-3\alpha^2}$, and these give the desired mean and second moment,
noting $\beta\ge \alpha^2$ since $E[X^2] \ge (E[X])^2$ so all the probabilities are between $0$ and $1$ and the "other point" is larger in magnitude than $\alpha$.  Thus
$$P \left[|X| > \frac{|\alpha|}{2} \right] \ge \frac{\alpha^2}{4\beta-3\alpha^2}$$
When $\alpha=0$, for some arbitrarily small $\delta$, let $X=+\sqrt{\frac \beta\delta}$ with probability $\frac{\delta}2$,  $X=-\sqrt{\frac \beta\delta}$ with probability $\frac{\delta}2$ and $X=0$ with probability $1 -{\delta}$.  So the lower bound in this situation  is $0$, as suggested by the earlier expression.
