Prove Standard deviation greater than or equal to Mean deviation Ho do we prove that the standard deviation is greater than or equal to the mean deviation about the arithmetic mean ?

$$
\sqrt\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n}\geq\frac{\sum_{i=1}^{n}|x_i-\bar{x}|}{n}
$$

and under what conditions we get the equality ?
I think i understand that it is because of the squaring in standard deviation which tends to give more weightage to the data far from the central tendency.
 A: Let $v=\left|\vec{x} - \bar{x}\vec{1}\right|$, where $|\cdot|$ is component-wise.
Then:
\begin{align}
\frac{1}{n^2}\left( \sum_i |x_i - \bar{x}| \right)^2
&= \frac{ \left(v\cdot\vec{1}\right)^2}{n^2}\\
&\leq \frac{(\vec{1}\cdot\vec{1})(v\cdot v)}{n^2}\\
&= \frac{1}{n}\sum_i|x_i - \bar{x}|^2
\end{align}
where we used the CS inequality for the second step. 
Now take the root of the first and last terms:
$$
\sqrt{\frac{1}{n}\sum_i|x_i - \bar{x}|^2\;} \,\geq \frac{1}{n}\sum_i |x_i - \bar{x}|
$$
A: The Cauchy Schwarz inequality 

$$\bigg(\sum_{i=1}^{n}a_{i}^2\bigg).\bigg(\sum_{i=1}^{n}b_{i}^2\bigg)\ge \bigg(\sum_{i=1}^{n}a_ib_i\bigg)^2
$$

taking $a_i=|x_i-\bar{x}|$ and $b_i=1/n$,
$$
\bigg(\sum_{i=1}^{n}|x_i-\bar{x}|^2\bigg).\bigg(\sum_{i=1}^{n}\tfrac{1}{n^2}\bigg)\ge \bigg(\sum_{i=1}^{n}|x_i-\bar{x}|.\tfrac{1}{n}\bigg)^2\\
\bigg(\sum_{i=1}^{n}(x_i-\bar{x})^2\bigg).\bigg(n.\tfrac{1}{n^2}\bigg)\ge \bigg(\frac{\sum_{i=1}^{n}|x_i-\bar{x}|}{{n}}\bigg)^2\\
\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}\ge \bigg(\frac{\sum_{i=1}^n|x_i-\bar{x}|}{n}\bigg)^2\\
\sqrt\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}\ge\frac{\sum_{i=1}^n|x_i-\bar{x}|}{n}\\
S.D\ge M.D
$$
