# To show that $\max\limits_{1\leq i\leq n}|x_i-\bar{x}|<\frac{n-1}{\sqrt n}\sqrt{\frac1{n-1}\smash[b]{\sum\limits_{i=1}^n}(x_i-\bar{x})^2}$

Let $x_1,\cdots,x_n \in \mathbb{R}$. Define $\bar{x}=\dfrac{1}{n}\sum\limits_{i=1}^n x_i$. I have to show the following. $$\max_{1 \leq i \leq n} |x_i-\bar{x}|<\frac{n-1}{\sqrt{n}}\sqrt{\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2}.$$ Unless all the $x_i$'s are equal or exactly $n-1$ $x_i$'s are equal.

This is equivalent as showing $$\max_{1 \leq i \leq n} (x_i-\bar{x})^2<\frac{(n-1)^2}{n}\sum_{i=1}^n(x_i-\bar{x})^2$$ By taking $y_i=x_i-\bar{x}$, we can reduce this to showing that $$\max_{1 \leq i \leq n} y_i^2<\frac{(n-1)^2}{n}\sum_{i=1}^n y_i^2, \quad\text{where }\sum_{i=1}^n y_i=0.$$ Any help regarding establishing the last inequality would be much appreciated. Thank you.

• This is Samuelson's inequality. There must have been a previous post or two. – StubbornAtom May 18 '18 at 6:31
• Thanks for giving me the name of the inequality. However, I'm not being able to find a quick proof of it. – pkxop May 18 '18 at 6:36
• Asked here before and a detailed post is here at CV. Or if you wish, this is Samuelson's original paper:jstor.org/stable/2285901?seq=1#page_scan_tab_contents. – StubbornAtom May 18 '18 at 6:37