How to prove the sum of squares is minimum?

Given $n$ nonnegative values. Their sum is $k$. $$ x_1 + x_2 + \cdots + x_n = k $$ The sum of their squares is defined as: $$ x_1^2 + x_2^2 + \cdots + x_n^2 $$

I think that the sum of squares is minimum when $x_1 = x_2 = \cdots = x_n$. But I can't figure out how to prove it. Can anybody help me on this? Thanks.

Solution 1:

HINT: By Cauchy-Schwarz we know

$$\left(\sum x_i y_i\right)^2 \leq \left(\sum x_i^2\right) \left(\sum y_i^2\right)$$

Take $y_i = 1$ for all $i$ to get a lower bound on $\sum x_i^2$. Then show that $x_i = \frac{k}{n}$ achieves this bound.

Solution 2:

Let $c = k/n$. Then, for all $(x_1,\ldots,x_n)$ such that $\sum_i x_i = k$, $$ \newcommand{\s}{\sum_{i=1}^n} \s x_i^2 = \s (c + x_i - c)^2 = c^2 n + \s (x_i - c)^2 \>, $$ since $2 \s c(x_i-c) = 0$. The right-hand side is obviously minimized by taking $x_i = c$ for all $i$ and so the result follows.