Show that $|\operatorname{median}(X) - \operatorname{mean}(x)| \leq \sigma_X$ Let $m$ denote median and $\bar{x}$ denote mean and $\sigma$ denote the standard deviation, I want to show that $|m - \bar{x}| \leq \sigma$. Since the LHS and RHS are both positive, we can prove $(m - \bar{x})^2 \leq \sigma^2$ instead.
Here is what I've attempted with my first approach:
\begin{align}
& |m - \bar{x}|^2 = |\bar{x} - m|^2 \\
= {} & \left|\sum_i (\frac{1}{n}x_i) - m\right|^2 \\
= {} & \left|\sum_i (\frac{1}{n}x_i) - \frac{n}{n}m\right|^2 \\
= {} & \left|\frac{1}{n}\sum_i x_i - m\right|^2 \\
= {} & \left(\frac{1}{n}\sum_i x_i - m\right)^2
\end{align}
Also,
$$
\sigma^2 = \frac{1}{n}\sum_i (x_i - \bar{x})^2
$$
I don't see an easy way to show that this quantity is $\geq$ than the previous quantity. Is this in the right direction? The standard deviation and mean are related, but it's not clear to me how the median relates to either one.
I'm not sure if this is relevant, but I also know that the minimizers for the following are the mean and median, respectively
$$
\bar{x} = \arg \min_y \sum_i (x_i - y)^2 \\
m = \arg \min_y \sum_i \left| x_i - y \right| \\
$$

This is easily obtained by convexity and Cauchy-Schwarz:
$$|\overline{x}-m| \le \frac 1n \sum |x_i -m| \le \frac 1n \sum |x_i -\overline x| \le \sqrt{\frac 1n \sum |x_i -\overline x|^2} = \sigma$$
