WakeUp wrote: Prove that the arithmetic mean $a$ of $x_1,\ldots ,x_n$ satisfies \[ (x_1-a)^2+\ldots +(x_n-a)^2\le (|x_1-a|+\ldots +|x_n-a|)^2\] Are you sure that the problem statement is true? It is obvious.

It is now fixed, apologies.

it's equal to x1,x2,...,xn satisfy x1+x2+x3+..+xn=0 , this way ,i used to meet this one !x1!+!x2!+...+!xn-1!>!xn! we are done =v=

WakeUp wrote: Prove that the arithmetic mean $a$ of $x_1,\ldots ,x_n$ satisfies \[ (x_1-a)^2+\ldots +(x_n-a)^2\le \frac{1}{2}(|x_1-a|+\ldots +|x_n-a|)^2\] Let $y_i = x_i - a$. We can see, that $\sum_{i=1}^n y_i = 0$. $|y_1|\sum_{i=1}^n |y_i| = y_1^2 + |y_1|\sum_{i=2}^n |y_i| \geq y_1^2 + |y_1||\sum_{i=2}^n y_i| = 2 y_1^2$. With $n - 1$ inequalities for $y_2, ..., y_n$, like for $y_1$. We can get our result! That's all! !