Problem: For integer $n \ge 2$,
let $x_1$, $x_2$, \dots, $x_n$ be real numbers satisfying
\[ x_1 + x_2 +  \dots + x_n = 0
  \quad\text{and}\quad x_1^2 + x_2^2 + \dots + x_n^2 = 1. \]
For each subset $A\subseteq\{1, 2, \dots, n\}$, define $S_A=\sum_{i\in A} x_i$.
(If $A$ is the empty set, then $S_A=0$.)
Prove that for any positive number $\lambda$,
the number of sets $A$ satisfying $S_A\geq\lambda$ is at most $2^{n-3}/\lambda^2$.
For which choices of $x_1$, $x_2$, \dots, $x_n$, $\lambda$ does equality hold?

\end{enumerate}

Solution: Let $\eps_i$ be a coin flip of $0$ or $1$.
Then we have
\begin{align*} \mathbb E[S_A^2]
  &= \mathbb E\left[ \left( \sum \eps_i x_i \right)^2 \right]
  = \sum_i \mathbb E[\eps_i^2] x_i^2
  + \sum_{i < j} \mathbb E[\eps_i\eps_j] 2x_i x_j \\
  &= \half \sum x_i^2 + \half \sum x_i x_j
  = \half + \half \sum_{i<j} x_i x_j
  = \half + \half \left( -\half \right) = \frac 14.
\end{align*}
In other words, $\sum_A S_A^2 = 2^{n-2}$.
Since can always pair $A$ with its complement, we conclude
\[ \sum_{S_A > 0} S_A^2 = 2^{n-3}. \]

Equality holds iff $S_A \in \{\pm \lambda, 0\}$ for every $A$.
This occurs when $x_1 = 1/\sqrt2$, $x_2 = -1/\sqrt2$,
$x_3 = \dots = 0$ (or permutations), and $\lambda = 1/\sqrt2$.