Proving $\frac{(a_1)^2+(a_2)^2+\cdots+(a_n)^2}{n}\ge \left(\frac{a_1+a_2+a_3+\cdots+a_n}{n}\right)^2 $ for positive real $a_i$

Question

$$\frac{(a_1)^2+(a_2)^2+\cdots+(a_n)^2}{n}\ge \left(\frac{a_1+a_2+a_3+\cdots+a_n}{n}\right)^2 $$ given that $(a_1), (a_2), \ldots, (a_n)$ are positive real numbers

I tried to prove this using AM GM but no luck there, so I tried to assume a function (then using derivative analysis hoping something happen but nothing).

Can anyone give me a hint????

Answer 1

By C-S $$\frac{\sum\limits_{i=1}^na_i^2}{n}=\frac{\sum\limits_{i=1}^n1^2\sum\limits_{i=1}^na_i^2}{n^2}\geq\frac{\left(\sum\limits_{i=1}^na_i\right)^2}{n^2}=\left(\frac{\sum\limits_{i=1}^na_i}{n}\right)^2.$$ Another way:

It's just Jensen for the convex function $f(x)=x^2.$

Another way: $$n\sum_{i=1}^na_i^2-\left(\sum_{i=1}^na_i\right)^2=\sum_{1\leq i<j\leq n}(a_i-a_j)^2\geq0.$$