Why isn't this expression returning true to being positive when it is clearly positive?

Question

Consider:

Expre10 = (B^2/C + 2 B + B^2/D + (B C)/D + (B D)/C) + 
   1 (C + D) - (A \[Beta] \[Sigma])/(B C D) (C + D);
Assuming[{A > 0, B > 0, C > 0, 
  D > 0, \[Beta] > 0, \[Sigma] > 0, (A \[Beta] \[Sigma])/(B C D) <=  
   1}, Simplify[Expre10 > 0]]

Returns the inequality:

B (B + C) (B + D) > A [Beta] [Sigma]

But we clearly see if (A \[Beta] \[Sigma])/(B C D) <= 1 holds then our inequality will always be true, so why am I getting the wrong output?

Answer 1

Simplify is at root an expression tree minimizer, equipped with some algebraic and logical transformations. As such, it may have the transformations needed to reach your goal, but its main goal is to apply transformations that result in smaller expression trees. It is also possible that it may not try the transformation needed to get to your goal. Functions whose purpose is to solve algebraic systems, such as Reduce, are generally more robust. They tend not to mind if all the cases to consider get quite complicated. Whereas Simplify is like a first-year university student who feels it's time to give up when things get complicated, Reduce is like a graduate student determined to impress their professor.

Expre10 = (B^2/C + 2 B + B^2/D + (B C)/D + (B D)/C) + 
     1 (C + D) - (A β σ)/(B C D) (C + D) /. C -> c /. 
   D -> d;
Assuming[{A > 0, B > 0, c > 0, 
   d > 0, β > 0, σ > 0,
   (A β σ)/(B c d) <= 1}, 
 Reduce[$Assumptions \[Implies] Expre10 > 0, {}, Reals]]

(*  True  *)

Specifying the domain Reals is key here, too.