For $n$ points on a plane, prove that there are at most $3n$ pairs of vertices with distance 1 
Question: Given $n$ points in a plane, the distance between any $2$ vertices is at least $1$. Prove there are at most $3n$ pairs of points with distance of exactly $1$.

I've seen this thread, which looks very very similar:  Given n points in the plane, such that the minimal euclidian distance is 1, show that there are at most 3n pairs of points with distance exactly 1
However it is about planar graphs - and we did not study what is a planar graph yet (and we won't in the future) Is there a way to solve it without using properties of planer graphs?
I did not understand how to start the proof without using planar graphs theory.. Thank you!

Sure, you can prove this without any graph theory at all.
Suppose there are more than $3n$ distances of length 1. This means more than 6 such distances per point (each distance is a distance between two points, hence it counts twice). This in turn means at least one point with at least 7 neighbors at distance 1, which themselves are at least 1 apart. And that just can't be done, regular hexagon is my witness.
So it goes.
