TITLE: Solution for sets of $(x_k-x_l)(x_l-x_m)(x_m-x_k)>0$
QUESTION [5 upvotes]: Given a set of inequalities like the following:
$$
(x_k-x_l)(x_l-x_m)(x_m-x_k)>0, 
$$
with $x_n\in\mathbb N_0$. These inequalities have solutions, when $\{x_k,x_l,x_m\}$ obeys a cyclic ordering like $\{0,1,2\}$, e.g. $(0-1)(1-2)(2-0)=2>0$.
How to show that a given set of inequalities has or does not have a solution? 
This is easy when all inequalities have independant variables. How entangled can these systems get before it gets impossible to solve them? 
EDIT: The question is in fact graph theorectically motivated: How to colour the edges of a $3$-regular simple graph (e.g. below) so that the colours of the edges that meet at a point always obey a cyclic clockwise ordering?
$\hskip1.5in$

REPLY [2 votes]: The graph-theoretical problem for cube can be simplified by taking the quotient of cube by a symmetry subgroup that 

preserves the orientation of cube's surface,
does not identify edges starting in any vertex.

Then you change equivalent numbers a bit and get a solution.
For example, first take a symmetry around the center of cube, see the left image for the quotient space, then take a rotation in the plane of the monitor by $180°$, see the right one.