*First (from last), the reason that I assumed without possible equality is that the side lengths of the triangle are not equal (unless it's degnerate in which case the result is still true).
*There's no need to consider all orderings (six of them), if I wrote instead the argument would be the same. That's what I mean by "without loss of generality".
Now, if you employ Opalg's additional line of reasoning in Post # 16, you get that all three are less than 1 in just one case, and the three variables appear symmetrically in the proof. Then you can use the without loss of generality to save yourself the trouble of writing the other cases.
Concerning equality: I don't see why isosceles triangles should be excluded from the proof. That's not a degenerate case. But not allowing equality in your inequalities leaves out the isosceles case, and for that matter the equilateral case. You could do those separately, I suppose. But Opalg's reasoning gets you there using the more general (in)equalities.
The multiplication is :
(1-a)(1-b)(1-c)>0 <=> abc +a+b+c <2 <=> (a+1)(b+1)(c+1)<4