$2$ out of $4$ points, each of distance at most $1$ apart, are at most $1/\sqrt2$ apart Given $4$ points in the plane such that any pair of them are a distance of at most $1$ apart, show that some pair of them must be of distance at most $1/\sqrt{2}$  apart.
I figured the solution might involve a pigeonhole argument, but as my geometry is rusty I'm having trouble turning the "distance of at most $1$" condition into a condition that bounds all $4$ points in a region that can then be subdivided into $3$ regions in which every point must be within $1/\sqrt{2}$. Or that might be entirely on the wrong track.

By eyeballfrog's comment, if there's an arrangement of four points such that all pairwise distances lie in the interval $(1/\sqrt{2}, 1]$, then by scaling, we can guarantee that two are a distance $1$ apart. Let these points be $A$ and $B$. The diagram shows circles of radius $1/\sqrt{2}$ and $1$ with centers $A$ and $B$. The remaining two points must be in the curvilinear regions $CDEF$ and $C'D'E'F'$, excluding the inner arcs of radius $1/\sqrt{2}$.
They can't be in different regions, as $ACBC'$ is a square with side-length $1/\sqrt{2}$, so $CC' = 1$, and between any two points in different regions there is a vertical distance of at least $CC'$.
But they can't be in the same region either. The triangle $ABE$ is equilateral, so $EC = \sqrt{3}/{2} - 1/2$. Applying the law of cosines to the triangle $ABD$ gives $$AD^2 = AB^2 + BD^2 - 2 AB \cdot BD \cos \angle ABD.$$ As $AB = BD = 1$ and $AD = 1/\sqrt{2}$, this gives $\cos \angle ABD = BG = 3/4$ and thus $DF = 1/2$ (because $DF \parallel AB$). Therefore, $CDEF$ can be contained in a rectangle with sides through $C, D, E, F$ and parallel to $CE$ and $DF$. This rectangle has diagonal length $\sqrt{CE^2 + DF^2} < 1/\sqrt{2}$, so all points in $CDEF$ (and, by symmetry, $C'D'E'F'$) are closer than $1/\sqrt{2}$.

