Sum of squares and the linear sum

Question

I'm wondering if I know for a fact

$$\sum_{i=1}^N x_i = k$$

Can I say anything about

$$\sum_{i=1}^N x_i^2 = ?$$

Answer 1

$$\sum\limits_{i=1}^N x_i^2 \ge \frac{k^2}{N}$$ but there is no upper bound in general, though there is if each $x_i\ge 0$.

As an example, let $x_i=z$ and $x_2=-z$ so $\sum\limits_{i=1}^2 x_i =0$ but $\sum\limits_{i=1}^2 x_i^2 = 2z^2$ which is unbounded above.

If you restrict $x_i$ to non-negative values then you can say (like Actually Fritz) that $\left(\sum\limits_{i=1}^N x_i\right)^2 \ge \sum\limits_{i=1}^N x_i^2$ by expanding the square of the sum, and so get an upper bound of $\sum\limits_{i=1}^N x_i^2 \le k^2$ for the sum of the squares, and equality iff all but one of the $x_i$ are $0$.

As a proof of the lower bound, you have $$\sum\limits_{i=1}^N \left(x_i-\frac kN\right)^2 \ge 0$$ $$\sum\limits_{i=1}^N x_i^2-2\sum\limits_{i=1}^N x_i \frac kN+ \sum\limits_{i=1}^N \left(\frac kN\right)^2 \ge 0$$ $$\sum\limits_{i=1}^N x_i^2-\frac {k^2}N \ge 0$$ and equality iff each $x_i =\frac{k}{N}$.

Answer 2

I'm afraid, there is not much you can say. Consider the simplest case of $x_1$ and $x_2$ $$ x_1 + x_2 = k. $$ For any $$ 2s^2 > k^2 $$ I can find $x_1$ and $x_2$ such that $$ \begin{aligned} x_1 + x_2 &= k,\\ x_1^2 + x_2^2 &= s^2. \end{aligned} $$ Proof. Let's do it constructive. Suppose $$ \begin{aligned} x_1 &= \frac{k-\sqrt{2s^2-k^2}}{2},\\ x_2 &= \frac{k+\sqrt{2s^2-k^2}}{2}. \end{aligned} $$ Obviously $$ x_1 + x_2 = \frac{k-\sqrt{2s^2-k^2}+k+\sqrt{2s^2-k^2}}{2} = \frac{2k}{2} = k $$ and $$ \begin{aligned} x_1^2 + x_2^2 &= \frac{k^2-2k\sqrt{2s^2-k^2}+(2s^2-k^2)+k^2+2k\sqrt{2s^2-k^2}+(2s^2-k^2)}{4} =\\ &= \frac{4s^2}{4} = s^2 \end{aligned} $$ QED.