Can the mean deviation about mean exceed the standard deviation for the Pareto distribution? Can the mean deviation about mean exceed the standard deviation for the Pareto distribution?
I just went through some books and found they are claiming that it cannot. 
How can I prove that? What is the underlying logic that confirms this conclusion?
 A: There's a well-known result, Jensen's inequality, which for our present purposes can be taken as:

if X is a random variable and $\varphi$ is a convex function, then $\varphi\left(\mathbb{E}\left[X\right]\right) \leq \mathbb{E}\left[\varphi(X)\right]$



*

*Population mean deviation and standard deviation
Consider the original variable to be $Y$, and let $X = |Y-\mu_Y|$. Further, let $\varphi(X)=X^2$. Then immediately by the above inequality, 
$$\left(\mathbb{E}\left[|Y-\mu_Y|\right]\right)^2 \leq \mathbb{E}\left[|Y-\mu_Y|^2\right]\quad\text{,}$$
and since both expectations are positive, 
$$\mathbb{E}\left[|Y-\mu_Y|\right] \leq \sqrt{\mathbb{E}\left[(Y-\mu_Y)^2\right]}\quad\text{,}$$
which is the required result. 

*Sample mean deviation (MD) and standard deviation, $s$
The above result applies to discrete distributions just as well as continuous ones, and so in samples, if we were to use an $n$ denominator for the standard deviation (i.e. $s_n$ instead of $s_{n-1}$) the same result clearly holds simply by applying the above result to the ECDF as if it were a CDF. 
Immediately we have, if $\underline x = (x_1, x_2, \ldots, x_n)$,
$$\text{MD}(\underline{x}) \leq s_n(\underline{x})$$. 
Since $s_{n-1} > s_n$, in that case it becomes a strict inequality.
(Alternatively, we could use one of the other forms of the inequality at the above link to get to the same result.)
As you see, we don't need to know it's Pareto at all. However, proving it specifically for the Pareto may be easier than proving Jensen's inequality in general, but probably not by as much as all that.
