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.

- Sub-case

and satisfying inequality - Sub-case

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.

**General case**

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~~ when LRC holds they are satisfied!!! )

Lets rewrite it again

if

It is not clear from this why is tight distance.