In other words, for a set of $n$ distinct points, with $d_{ij}$ being the distance between point $i$ and point $j$, is the $n \times n$ matrix $M = [e^{-d_{ij}}]$ positive semidefinite?
I've written some code to look for counterexamples and haven't found any, so I'm quite sure it's true, but I can't find a proof online.
So far, I've found some interesting facts about the matrix $M$, but I don't know if they lead anywhere. $$M^2_{ij} \leq n M_{ij}$$ $$|M| = \hspace{-1.2cm} \sum_{g \in \textrm{directed graphs over $n$ points, where each point has out-degree and in-degree 1}} \hspace{-4cm} (-1)^{\textrm{# of loops with even # of points}} e^{-\textrm{sum of lengths of edges in }g}$$
Also, I think I can prove the result given a proof of the following:
For a bipartite graph, if you take the softmin over $(d_{ij})_{i \in [n], j \in [n], i \textrm{ and } j \textrm{ on same side}}$ that is bigger than the softmin over $(d_{ij})_{i \in [n], j \in [n], i \textrm{ and } j \textrm{ on opposite sides}}$. (Note that distances between distinct points are being double-counted, but distances between a point and itself are not.)