Is mean absolute deviation smaller than standard deviation for $n\ge 3$? I want to compare the mean absolute deviation with standard deviation in general case with this definition:
$$MAD = \frac{1}{n-1}\sum_1^n|x_i - \mu|, \qquad SD = \sqrt{\frac{\sum_1^n(x_i-\mu)^2}{n-1}}$$
where $\mu =\frac{1}{n}\sum_1^n x_i$.
Is it true that $MAD \le SD$ for every $\{x_i\}^n_1$?
It's false for $n=2$, becouse $x+y \ge \sqrt{x^2+y^2}$, for every $x, y \ge 0$.
It's easy to show that:
$$MAD \le \sqrt{\frac{n}{n-1}} \times SD$$
 A: Here is a more mathematical approach.  Firstly, it's probably true that by a change of variables, one can assume that the mean is zero. Certainly from the point of view of finding a counter example, this is acceptable.  So, setting $ \mu = 0$, squaring both sides of the proposed inequality and multiplying through  by (n-1) one is left with the proposed inequality - 
$ (\sum_{i=1}^{i=n}|x_i|)^2 \leq (n-1)(\sum_{i=1}^{i=n}|x_i|^2))$ 
This looks fishy.  (n-1) isn't enough to make up for all the $|x_i| |x_j|$ terms . Particularly if all the $x_i$ are the same in absolute value.  My first guess was n=4 and $ x_1 = x_2 =1, x_3=x_4 = -1$.  This leads to $\frac{4}{3} \leq \sqrt{\frac{4}{3}}$ . 
I would think that this sort of thing is well known to people interested in inequalities. 
A: No, in general this is not true.
A simple way to look at this is to simulate. I typically hack together an infinite loop that stops if it finds a counterexample. If it runs for a long time, I start thinking about whether the claim might be true. In the present case, my R code looks like this:
while ( TRUE ) {
    xx <- runif(3)
    mad <- sum(abs(xx-mean(xx)))/(length(xx)-1)
    sd <- sqrt(sum((xx-mean(xx))^2)/(length(xx)-1))
    if ( mad > sd ) break
}
xx

It yields this counterexample:
[1] 0.7852480 0.0760231 0.8295893

