Where $$s = \sqrt{ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2}$$ and $$ M = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|$$
I came up with a sketchy proof for the case of $2$ values, but I would like a way to generalize (my "proof" unfortunately doesn't, as far as I can tell).
- Proof for $2$ values (I would appreciate feedback on this as well):
$$\frac{1}{\sqrt{2}} \sqrt{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2} \geq \frac{1}{2} (|x_1- \bar{x}| + |x_2- \bar{x}|)$$
Now let $|x_1- \bar{x}| = a$ and $|x_2- \bar{x}| = b$ be the $2$ legs of a right triangle and $\sqrt{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2} = c$ its hypothenuse.
And let $\theta$ be the angle between $c$ and either $a$ or $b$.
Then $(\sin{\theta} + \cos{\theta}) = \frac{a}{c} + \frac{b}{c} = \frac{a+b}{c} = \frac{\frac{1}{2} (|x_1- \bar{x}| + |x_2- \bar{x}|)}{\frac{1}{\sqrt{2}} \sqrt{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2}}$
so that $$ \sqrt{2} \geq (\sin{\theta} + \cos{\theta})$$
And we know that $\max(\sin{\theta} + \cos{\theta}) = \sqrt{2}$. QED?
I have no idea how to prove the general case, though.
Let $y_i = x_i - \bar{x}$ then by Cauchy-Schwarz we have the following:
$\sum_{i=1}^n |y_i| \frac{1}{n} \leq \|y\|_2 \sqrt{\sum_{i=1}^n \frac{1}{n^2}} = \sqrt{\frac{1}{n}\sum_{i=1}^n |x_i - \bar{x}|^2}$