It's not a full explanation (i may come back to it depending on the comments), but here's an intuitive derivation:
Denoting $\mu$ the mean ($\neq$ average), $m$ the median, $\sigma$ one standard deviation, $M$ the mode, $sgn()$ the sign function and $X$ the (random) dataset.
It's well known that $|\mu-m|\leq\sigma$. Certainly, Chebyshev's inequality must imply that $|\mu-M|<3\sigma$. Finally, it's easy to show that $sgn(\mu-M)=sgn(\mu-m)$ (with the convention that if $sgn(\mu-m)==0$, then both sides equal 1).
All these hold for any unimodal distribution (regardless of skew).
EDIT: this is a frequent textbook exercise:
&\leq& E|X-m| \
&\leq& E|X-\mu| \
&=& E\sqrt((X-\mu)^2) \
&\leq& \sqrt(E(X-\mu)^2) \
The first equality derives from the definition of the mean, the third comes about because the mean is the unique minimiser (among all $c$'s) of $E|X-c|$ and the fourth from Jensen's inequality (i.e. the definition of a convex function).
Chebyshef's inequality states that: