How can I prove that the mean of squared data points is greater than the square of the mean of the data points? The inequality I'm trying to prove is
$$\frac{1}{n}\sum_{i=1}^nx_i^2 \geq \frac{1}{n^2}\left(\sum_{i=1}^nx_i\right)^2.$$
I tried simply expanding out the RHS but I can't seem to count the terms properly. I can reduce the inequality to
$$(n-1)\sum_{i=1}^nx_i^2 - 2\sum_{i\neq j}x_ix_j \geq 0$$
but there appear to be $2(n-1)!$ terms in this $i\neq j$ sum and only $n(n-1)$ terms in the sum of the squares, which doesn't seem right to me.
I've also "assumed" that $x_1 \leq\dots\leq x_n$ which allows me to set up the inequalities $x_1^2 \leq x_1x_2 \leq x_2^2$ etc. but my counting is letting me down.
Is there an easier way to do this?
 A: This is Jensen's inequality applied to the set $(x_1, \ldots, x_n)$ using the convex function $f(x) = x^2$.
It is also directly provable using the following elementary argument:  define $\bar x = \frac{1}{n} \sum_{i=1}^n x_i$ to be the sample mean.  Now we have 
$$ \begin{align*} \sum_{i=1}^n x_i^2 
&= \sum_{i=1}^n (x_i - \bar x + \bar x)^2 \\
&= \sum_{i=1}^n \left( (x_i - \bar x)^2 + 2\bar x (x_i - \bar x) + \bar x^2 \right) \\
&= \sum_{i=1}^n (x_i - \bar x)^2 + 2 \bar x \left( -n \bar x + \sum_{i=1}^n x_i \right) + n \bar x^2 \\
&=  n \bar x^2 + \sum_{i=1}^n (x_i - \bar x)^2  \\
&\ge n \bar x^2.\end{align*}$$
The second equality comes from expanding the square; the third comes from distributing the sum over the three terms in the summand; the fourth equality is a result of the fact that $n \bar x = \sum_{i=1}^n x_i$, thus the middle term equals zero; and the final inequality is a consequence of the fact that no real square is negative, thus the sum is bounded below by $0$.
Now all that remains is to divide by $n$ to obtain $$\frac{1}{n} \sum_{i=1}^n x_i^2 \ge \bar x^2 = \left(\frac{1}{n} \sum_{i=1}^n x_i \right)^2,$$ as claimed.
