Force Mathematica to simplify a function to avoid infinities

Question

I am doing some evaluations which return functions of three variables similar to the following:

Q[x_,y_,z_] = 1/2 (1 - (-1 + 2 y z + 2 x (y + z)) Abs[(
 1 + 4 x^2 + 2 y (-2 + 2 y + z) + 2 x (-2 + 3 y + z))/(-1 + 
  2 y z + 2 x (y + z))] + (x y + (x + y) z) Abs[
 2 - (4 x y)/(y z + x (y + z))])

I want to create a RegionPlot3D of $Q-\frac{1}{2}>0$ in the region with $x,y,z \in[0,1]$, along with another constraint $x+y+z\leq1$, but notice that this function $Q$, as it is currently written, will encounter some infinity problems. For example, the point $x=y=z=0$, will lead to an infinity inside the second Abs term, due to the xy+yz+zx in the denominator. In practice, however, this term is premultiplied with xy+yz+zx, so that should not actually cause an issue. For this simple example, I can just manually simplify both of the Abs terms by taking the premultiplying factor inside the Abs and cancel the corresponding term (also making sure I keep a Sign[] of the term moved inside the absolute value). As an example, the first absolute value term should be

(-1 + 2 y z + 2 x (y + z)) Abs[(1 + 4 x^2 + 2 y (-2 + 2 y + z) 
+ 2 x (-2 + 3 y + z))/(-1 + 2 y z + 2 x (y + z))] 
= Sign[-1 + 2 y z + 2 x (y + z)] Abs[1 + 4 x^2 + 2 y (-2 + 2 y + z) + 2 x (-2 + 3 y + z)] 

With these simplification, when I make a RegionPlot3D I get no error. However, in practice the functions I am looking at have many terms of this form and it is becoming impractical to do these simplifications by hand. So, my question is, how can I tell Mathematica to do that? I have tried both Simplify and FullSimplify and both fail to make that simplification. Thanks!

Answer 1

You may construct a list of exclusions

 Plot[... , Exclusion-> ...] 

e.g. for infinities by zero denominators by

     (FindInstance[0 == #, Union@Cases[{#}, _Symbol, \[Infinity]]] &) /@ 
     (1/ Cases[ Q[x, y, z], _^(_?Negative), \[Infinity]]) 

      {{{x -> -(3/2), y -> 1, z -> -4}}, {{x -> 0, y -> 0, z -> 0}}}

Other singular terms can be found by similar methods replacing the pattern of a negative power

   _^(_?Negative)  

by another special pattern and determining its singularities by specialized methods.

Answer 2

This can be done as follows.

red = Reduce[(1 - (-1 + 2*y*z + 2*x*(y + z))*
      RealAbs[(1 + 4*x^2 + 2*y*(-2 + 2*y + z) + 
          2*x*(-2 + 3*y + z))/(-1 + 2*y*z + 2*x*(y + z))] + (x*
         y + (x + y)*z)*RealAbs[2 - (4*x*y)/(y*z + x*(y + z))])/
   2 (1 - (-1 + 2*y*z + 2*x*(y + z))*
      RealAbs[(1 + 4*x^2 + 2*y*(-2 + 2*y + z) + 
          2*x*(-2 + 3*y + z))/(-1 + 2*y*z + 2*x*(y + z))] + (x*
         y + (x + y)*z)*RealAbs[2 - (4*x*y)/(y*z + x*(y + z))])/
   2 > 1/2 && x >= 0 && x <= 1 && y >= 0 && y <= 1 && z >= 0 && 
z <= 1];
RegionPlot3D[red, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, PlotPoints -> 50]