Let $S$ be a finite set of unit vectors in $d$-dimensional complex Hilbert space $l_2^d$.
I believe (from numerical experiment) that $\frac{1}{\lvert S \rvert^2} \sum_{x,y \in S} \lvert \langle x,y \rangle \rvert^2 \geq \frac1d$. In other words, if I choose two vectors from $S$ at random and take the squared absolute value of their inner product, the expectation is at least $\frac1d$, for any initial choice of $S$.
This feels like it should have a simple proof but it has eluded me.
Edited to add: a more general, but actually equivalent, statement is the following. Given an arbitrary probability distribution on the unit sphere of $l_2^d(\mathbb{C})$, if $x$ and $y$ are chosen independently from this distribution, then $\mathbf{E}(\lvert \langle x, y \rangle \rvert^2) \geq \frac1d$.
Maybe there is a simpler argument, but I came up with the following:
Given a probability distribution $P(x)$ on the unit sphere we can consider the matrix
$$ A = \int dx\, P(x)xx^\dagger. $$ This is readily seen to satisfy $$ A^\dagger=A,\qquad \operatorname{tr}A=1,\qquad y^\dagger A y\ge 0\quad \forall y\in \mathbb{C}^d, $$ i.e. it can be diagonalized with postive eigenvalues $\lambda_i$ that sum to 1. We can then compute the trace of the square $$ \operatorname{tr} A^2 = \int dx\, dy\,P(x)P(y)\operatorname{tr}(xx^\dagger y y^\dagger) = \int dx\, dy\,P(x)P(y)\, |x^\dagger y|^2=\mathbf{E}(\lvert \langle x, y \rangle \rvert^2), $$ which is the expectation value of interest. As this equals the trace of $A^2$ we can equally consider the sum of the squared eigenvalues, which can be bound from below by the Cauchy–Schwarz inequality $$ \mathbf{E}(\lvert \langle x, y \rangle \rvert^2)=\sum_{i=1}^d\lambda_i^2\ge \frac{1}{d}\left(\sum_{i=1}^d \lambda_i\right)^2=\frac{1}{d}. $$