$n\geq 3$ points lie in three-dimensional space. What is the largest $c(n)$ such that there always exists a point for which the ratio between the distance to the farthest point from it and the distance to the second farthest point from it is at most $c(n)$?
In the case where $n=3$ and all three points are on a line, we can suppose that the distance between the first and second point is $a$, and between the second and third point is $b$, where $a\leq b$. The ratios from each of the three points are $\frac{a+b}{a},\frac{b}{a},\frac{a+b}{b}$. Among these ratios, $\frac{a+b}{a}$ is the highest, so we may ignore it. The remaining two ratios are equalized when $\frac{b}{a}=\frac{\sqrt{5}+1}{2}$.
$c(n)->1$, based on my data at BiggestLittlePolyhedron. For the $n$ points you can normalize the points so that the largest distance is $1$. The solutions I have there maximize the volume. Your problem minimizes the second largest distance.
This problem might be better as finding the best configuration for a particular $n$.
With the maximal distance being a unit distance, here are my best results for the minimal second largest distance for $n=8,9,16$.