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?

  • 1
    $\begingroup$ Use Cauchy Schwarz inequality on $x_i,1$ $\endgroup$
    – user379195
    Commented May 9, 2017 at 0:14
  • $\begingroup$ Yes the required inequality or simply, $\mathrm{Var}(x_i)\ge0$ easily follows from the C/S inequality. $\endgroup$ Commented May 9, 2017 at 9:50

1 Answer 1


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.

  • $\begingroup$ Thanks a lot. Seems I missed the "add zero" trick. $\endgroup$ Commented May 9, 2017 at 0:39
  • 2
    $\begingroup$ The key to the elementary solution is that we are separating out the total sum of squares into a main component (the square of the means), plus a residual component (the sum of squared deviations from the mean). This is a theme that you will encounter repeatedly in the theory of linear models. $\endgroup$
    – heropup
    Commented May 9, 2017 at 13:41

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