We consider (LRC) .
Conjecture Suppose runners having distinct constant speeds start at a common point (origin) and run laps on a circular track with circumference 1. Then, there is a time when no runner is closer than from the origin.
W.l.o.g. we can assume .
One can formulate LRC as follows. Suppose is the number of whole laps (including 0) runner k passed on a track, than and .
Case n=1, two runners is trivial. At time runner 1 is exactly distance ().
Case n=2, three runners is a special case.
We start with 4 inequalities
Since we can eliminate it if all combination of left parts on the left columns are smaller than any right parts in the right column of the table. e.g.
The first and the last inequality are trivially correct. From the second and third inequality we would like to express
In other words,
There are 2 sub-cases.
and satisfying inequality
In this case lead to (remeber that )
Case n=3, four runners is an illustration for a general case.
We start with 6 inequalities
Let’s express in terms of
Rewriting it we obtain
since In other words, satisfying inequality.
Now, let express in terms of and
We can express with inequalities obtained earlier
Collecting terms we obtain
We can see that first inequality is always stronger that the second one (meaning that if satisfies first inequalities it will satisfy second inequalities). But the first inequalities are the same as previous inequalities for with relabelling. Therefore, satisfying initial inequalities and LRC holds.
We start with inequalities
Now we expressing in terms of and in terms of and
Now we can substitute first inequalities into the second ones.
Rearranging terms we obtain
Compare it to
The second inequality is always stronger.
Therefore, we left with the set of inequalities (when they are satisfied LRC holds )
Lets rewrite it again
It is not clear from this why is tight distance.