Suppose it is established that \(n\) runner running about a unit circle with arbitrary constant speeds, \(v_i\), one just experienced a moment of loneliness, and is at least \(\cfrac{1}{n}\) away from the closest runner.
We add the next \((n+1)\) runner, with relative speed zero, where all other runners take reference, right in between and achieve a least distance of \(\cfrac{1}{2n}\). This \((n+1)\) runner is at the origin zero (starting point). The added \((n+1)\) runner still has arbitrary speed, distinct and constant.
This is the trivial case of \(n=3\).
Since \(\cfrac{1}{2n}\lt\cfrac{1}{(n+1)}\)
for \(n\gt 2\)
Adding the next \((n+1)\) runner becomes trivial, and with the trivial case of \(n=3\), by induction the conjecture is true. The constraint is still a least distance of \(\cfrac{1}{n}\), it has not been changed to \(\cfrac{1}{2n}\).
What? Using relative speeds is not suppose to change any aspect of the analogy.
Happy?
Note: All speeds changed by a constant, but the transition of time has no bearing here. The origin is rotated by a constant. The circle is rotating at the reference speed, speed of the \((n+1)\) runner. Starting time is later for all.
