I found that the easiest way to do this is to approach the problem backwards. That is, think of some promising values for x, y and z, and see which values of w they work for.
For example, let and take , and . Then and the equation (*) becomes . The greatest of the numbers x,y and z is x, namely n(n+1). So that choice will work for all numbers w in the interval .
But each of those intervals overlaps with the next one, since . So taken together, those intervals cover all the natural numbers, except for those that lie before the start of the first interval. That is the interval corresponding to n=2, which covers the numbers . So we are left with having to deal with w=1,2,3,4 and 5.
For w=4 we can take x=2, y=3 and z=1. But now we can make use of the information that w is of the form uv (where u and v have no common divisor). I think that the question probably also wants us to assume that neither u nor v is equal to 1. If so, then the numbers 1,2,3 and 5 are not of the form uv so we don't have to worry about them anyway, and the proof is complete.
Edit. You may want to require that x, y and z should also be different from 1. In that case, take , for n=2,3,4,..., and use a similar line of reasoning.