Maximum and minimum distance of two points Consider six distinct points in a plane. Let $m$ and $M$ denote respectively the minimum and the maximum distance between any pair of points.
Show that $M/m \geqslant \sqrt{3}$.
 A: Assume otherwise.
Pick two points $A,B$ at distance $M$ (The small dots in the following image)

Then all other points are in the closed lens-shaped area bounded by the  circles of radius $M$ around $A$ and $B$ (big circles in above image). In particular, they must be in the ten blue triangles shown. 
As the blue triangles have side length $\frac M{\sqrt 3}$, each of them can contain at most one of the points (including on the boundary).
After taking away those already occupied by $A$ and B$, there are only four blue triangles left. We conclude that each of the top two and each of the bottom two blue triangles contains one of the four remaining points.
Next consider the red lines: The shape bounded from above by the top red line and from below by the two small circles has diameter $<\frac M{\sqrt 3}$, hence can contain at most one point.
We conclude that one of the points must be above the top red line.
By the same argument, one of the points must be below the bottom red line.
But then these two points are more than $M$ apart - contradiction!
