Quantifying how crowded points are in $d$ dimensional cube Let $x_1, \cdots, x_n$ be distinct points in the $d$ dimensional cube $[-1,1]^d$. What is a lower bound on the quantity:
$$ \sum_{1 \le j < k \le n} \frac{1}{||x_k-x_j||}$$
where $|| \cdot ||$ is the Euclidean norm.
An asymptotic answer for large $n$ is also fine and bounds on any quantity that looks similar, such as replacing the norm with norm squared, is also welcomed. 
My motivation is a problem that appeared in the book The Cauchy-Schwarz Master Class which proved the following:

If $-1 \le x_1 < x_2 < \cdots < x_n \le 1$ then $$  \sum_{1 \le j < k \le n} \frac{1}{x_k-x_j}  \ge \frac{n^2 \log n}8.$$


This is an answer only up to $n = \exp(ad)$ for some $a \in \theta(1)$, but may be enough to get started: There are as many as $n = \exp(ad)$ points $y_1,\ldots, y_n$ in $\{-1,1\}^d$ that satisfy the following:
$||y_i-y_j||_1 \in \theta(d)$ for each $i \not = j$, where $|| \cdot ||_1$ denotes the Manhattan metric.
 [Google error-correcting codes]
Then for each $i \not = j$, the following holds:
$$\frac{1}{||y_i-y_j||_2} \leq \frac{1}{c\sqrt{d}}$$
for some $c \in \Omega(1)$, which implies the following:

Inequality 1: $$\sum_{1 \le i < j \le n} \frac{1}{||y_i-y_j||_2} \leq \frac{n^2}{2c\sqrt{d}}$$

However. not that for any distinct $x,y$, the following holds:

Inequality 2: $$\frac{1}{||x-y||_2} \ge \frac{1}{\sqrt{d}} $$

So Inequalities 1 and 2 together imply that for $y_1,\ldots, y_n$ as above,
$\sum_{1 \le i < j \le n} \frac{1}{||y_i-y_j||_2}$ is no more than a constant  multiple larger than $\sum_{1 \le i < j \le n} \frac{1}{||x_i-x_j||_2}$ for any other $x_1,\ldots, x_n \in [-1,1]^d$.
