Mean, Median and Standard Deviation

We will solve for the general case. Using the following well-known optimality property of the median $m$ : $\mathbb{E}|X-m| \leq \mathbb{E}|X-t|, \hspace{10pt} \forall t \in \mathbb{R}.$ In particular, putting $t=\mu=\mathbb{E}(X)$ , we get $|\mu-m|\equiv |\mathbb{E}(X)-m|\stackrel{(a)}{\leq} \mathbb{E}|X-m| \leq \mathbb{E}|X-\mu| \stackrel{(b)}{\leq} \sqrt{\mathbb{E}(|X-\mu|)^2}\equiv \sigma ,$ where in (a) and (b), we have used the Jensen's inequality with the convex functions $x\to |x|$ and $x\to |x|^2$ respectively. This implies that $\mu-\sigma \leq m \leq \mu+\sigma.$ A discrete probability distribution which puts a proability mass of $0.5$ at the points $\mu+\sigma$ and $\mu-\sigma$ shows that the bound above is indeed tight. $\blacksquare$