TITLE: Prove that if $\sum_{i=1}^n x_i=0$ and $\sum_{i=1}^n x_i^2=1$ there exist $i,j$ such that $x_ix_j\leq -{1\over n}$
QUESTION [6 upvotes]: Problem
Let $x_1, x_2, \ldots, x_n$ be reals such that
  \begin{align}
\sum_{i=1}^n x_i=0
\qquad \text{ and } \qquad
\sum_{i=1}^n x_i^2=1 .
\end{align}
  Prove that there exist $i,j$ such that $x_ix_j\leq -{1\over n}$.

my attempt
I found that if there exist $x_i,x_j$ such that $$x_i\geq {1\over\sqrt{n}}, x_j\leq -{1\over \sqrt{n}} $$
then the problem solved, but I can't prove this.

REPLY [6 votes]: This is basically a rewrite of the answer of @user1551 in more elementary terms. Without loss of generality we can assume that $x_1$ is minimal in $\{x_1,\ldots,x_n\}$. Then also some other coordinate is strictly greater than $x_1$. Let $c_k = x_1x_k+\tfrac1n$. Then $\sum_k c_k = 1$ and $\sum_k c_k x_k=x_1$. Not all $c_k$ can be positive, otherwise this convex combination would be greater than the minimum $x_1$.
This can also be seen by rewriting as follows: $$x_1=\sum_k c_k x_k = \sum_k c_k(x_1 + x_k-x_1) = x_1 + \sum_k c_k (x_k-x_1).$$ So either all summands $c_k(x_k-x_1)$ are zero or at least one of these is negative. Both cases imply that $c_k\leq0$ for some $k$ (note that $x_k-x_1>0$ for at least one index $k$).