Denote $\mu$ the mean ($\neq$ average), $m$ the median, $\sigma$ the standard deviation and $M$ the mode. Finally, let $X$ be the sample, a realization of a continuous unimodal distribution $F$ for which the first two moments exit.

It's well known that 

$$|\mu-m|\leq\sigma\quad (1)$$

This is a frequent textbook exercise:  

\begin{eqnarray}
|\mu-m| &=&    |E(X-m)|          \\
        &\leq& E|X-m|            \\
        &\leq& E|X-\mu|          \\
        &=&    E\sqrt{(X-\mu)^2} \\
        &\leq& \sqrt{E(X-\mu)^2} \\
        &=&    \sigma
\end{eqnarray}
The first equality derives from the definition of the mean, the third comes about because the median 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). Actually, this inequality can be made tighter. In fact, for any $F$, satisfying the conditions above, it can be shown [3] that

$$(2)\;\;\;|m-\mu|\leq \sqrt{0.6}\sigma$$


Even though it is in general not true ([Abadir, 2005](http://www.jstor.org/stable/3533476)) that any unimodal distribution must satisfy either one of
 $$M\leq m\leq\mu\textit{ or }M\geq m\geq \mu\quad (2)$$ 
it can still be shown that the inequality 

$$|\mu-M|\leq\sqrt{3}\sigma\quad (3)$$

holds for any unimodal, square integrable distribution (regardless of skew). This is proven formally in [Johnson and  Rogers (1951)](http://www.jstor.org/stable/pdfplus/2236630.pdf?acceptTC=true&jpdConfirm=true) though the proof depends on many auxiliary lemma's that are hard to fit here. Go see the original paper.


A sufficient condition for a distribution $F$ to satisfy $\mu\leq m\leq M$ is given in [2]. If $F$:

$$(4)\;\;\;F(m−x)+F(m+x)\geq 1 \text{ for all }x$$

then $\mu\leq m\leq M$. Furthermore, if $\mu\neq M$, then the inequality is strict. The Pearson Type I to XII distributions are one example of family of distributions satisfying $(4)$ [4]. 

Now assuming that $(4)$ holds strictly and w.l.o.g. that $\sigma=1$, 
 $3(m-\mu)\in(0,3\sqrt{0.6}]$ and $M-\mu\in(m-\mu,\sqrt{3}]$, since these two ranges overlap, it's certainly possible to find distributions for which the assertion is true (e.g. for when $0<m-\mu<\frac{\sqrt{3}}{3\sqrt{0.6}}<\sigma=1$).



 - [0]: The Moment Problem for Unimodal Distributions. 
N. L. Johnson and C. A. Rogers. The Annals of Mathematical Statistics, Vol. 22, No. 3 (Sep., 1951), pp. 433-439
 - [1]: The Mean-Median-Mode Inequality: Counterexamples
Karim M. Abadir
Econometric Theory, Vol. 21, No. 2 (Apr., 2005), pp. 477-482
 - [2]: W. R. van Zwet, Mean, median, mode II, Statist. Neerlandica, 33 (1979), pp. 1--5.
 - [3]: The Mean, Median, and Mode of Unimodal Distributions:A Characterization. Theory Probab. Appl., 41(2), 210–223.
 - [4]: Some Remarks On The Mean, Median, Mode And Skewness. Michikazu Sato. Australian Journal of Statistics. Volume 39, Issue 2, pages 219–224, June 1997