Would this proof strategy work for proving the lonely runner conjecture? The problem is the lonely runner conjecture. This conjecture states that if  $k$ runners begin running down a circle of unit circumference with random speeds, it will always the case that all runners become lonely, that is, it will always be the case that each runner is separated by all other runners with at least a distance $1 / k$.
Basically, my strategy is to prove a closely related conjecture using proof by induction. This conjecture is the equal distance runner conjecture (EDRC): At some time $T_k$, for  $k$ runners with random speeds running around a circle with circumference 1, the runners all become separated by distance $1 / k$ at the same time. This implies they all become lonely at time $T_k$.
The key to proving this conjecture is to find a function $F$ that is maximized when all runners are equally far apart.  This is done with Lagrange multipliers. This function $F$ is simply the area of the polygon formed by connecting the runners together. For any $k$, $F$ reaches a maximum when $\vartheta$ = 2 $\varPi $ / $k$, where $\vartheta$ is the angle between every runner.
The induction part comes in by assuming that there exists a time $T_k$ for which $F_k$ is maximized. The base case is very easy. We then show that there exists $\varepsilon> 0$ for which $T_k + \varepsilon$ maximizes $F_{k+1}$. This means there is a future time for which all $k + 1$ runners are lonely, since they're all seperated by distance  $1 / (k +1)$.
To prove that $\varepsilon$ exists, we have to consider a fairly complicated infinite series and show that there exists values of epsilon for which the series equals zero. The infinite series is the sum of two other convergent series (this has been proven), so the sum is convergent.
Once we show that $\varepsilon > 0$ exists, this means that $T_k$ + $\varepsilon $ maximizes $F_{k + 1}$, where $F_{k + 1}$ is the area of the polygon made by $k + 1$ runners. This implies that the $k + 1$ runners are equally far apart (that is, a distance $1 /(k+1)$ apart), which implies they all become lonely at time $T_k + \varepsilon $. This implies the LRC.
 A: The equidistant runner conjecture is false for $3$ runners. For an explicit counterexample, look at a unit circle (radius $1$) with three runner speeds $1,2$, and $4$. This has the advantage of completely resetting at time $2\pi$, since then all runners are at the starting line again. It's not too hard to explicitly check this with a computer.
As is pointed out by r.e.s. in a comment, we can verify this elegantly by hand as well. Nothing changes if we subtract a constant speed from each runner. So we might consider the equivalent scenario where the speeds are $0, 1$, and $3$. (The lonely runner/stander conjecture, perhaps). Then to be equidistant, the runners need to be at 1/3 and 2/3 of the way around the circle. But when the speed $1$ runner is at either $1/3$ or $2/3$ of the way around the track, the speed $3$ runner is at $0$, consoling the lazy sitter. So there is never a case when these three runners are equidistant.
Alternately, you can examine this visualization of this case (which I found on reddit's r/lonelyrunners).

So no, this is not a good proof strategy for proving the lonely runner conjecture.
