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.

• The Cauchy–Schwarz inequality will do. – Chappers Jul 20 '17 at 18:18
• @Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it. – ss1729 Jul 20 '17 at 18:41

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: $$\sqrt{\frac{1}{n}\sum_i|x_i - \bar{x}|^2\;} \,\geq \frac{1}{n}\sum_i |x_i - \bar{x}|$$
• How does $\left|\vec{x} - \bar{x}\vec{1}\right|= \sum_i \left|x_i - \bar{x} \right|$ ? – ThePassenger Jul 21 '17 at 8:22
$$\bigg(\sum_{i=1}^{n}a_{i}^2\bigg).\bigg(\sum_{i=1}^{n}b_{i}^2\bigg)\ge \bigg(\sum_{i=1}^{n}a_ib_i\bigg)^2$$
taking $$a_i=|x_i-\bar{x}|$$ and $$b_i=1/n$$,
$$\bigg(\sum_{i=1}^{n}|x_i-\bar{x}|^2\bigg).\bigg(\sum_{i=1}^{n}\tfrac{1}{n^2}\bigg)\ge \bigg(\sum_{i=1}^{n}|x_i-\bar{x}|.\tfrac{1}{n}\bigg)^2\\ \bigg(\sum_{i=1}^{n}(x_i-\bar{x})^2\bigg).\bigg(n.\tfrac{1}{n^2}\bigg)\ge \bigg(\frac{\sum_{i=1}^{n}|x_i-\bar{x}|}{{n}}\bigg)^2\\ \frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}\ge \bigg(\frac{\sum_{i=1}^n|x_i-\bar{x}|}{n}\bigg)^2\\ \sqrt\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}\ge\frac{\sum_{i=1}^n|x_i-\bar{x}|}{n}\\ S.D\ge M.D$$