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Judge the sign of a multi parameter equality Using 'Simplify'

I'm not sure this method is available for judging the sign of a multi parameter equality using the function "Simplify" in Mathematica.

But in the official document, the "Simplify" could "use assumptions to prove inequalities".

So I tried this method to judge my complex equation (it's other formula's first derivative and I need to judge it positive or negative)

And I posted my code here, please help me why it can not obtain the outcome of "True of False".

Simplify[1 + (
   b^m E^((qL \[Rho])/\[Lambda]) m p^(-1 + m)
     RqL (E^(b/\[Lambda]) + 
      E^((qL \[Rho])/\[Lambda]) (-1 + \[Alpha]) - 
      E^((b^m qH)/((b^m + p^m) \[Lambda])) \[Alpha]))/((b^m + p^
      m)^2 (-E^(((2 b)/\[Lambda])) + E^((
      b + (b^m qH)/(b^m + p^m))/\[Lambda]) + 
      E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (-1 + \[Alpha]) - 
      E^((2 qL \[Rho])/\[Lambda]) (-1 + \[Alpha]))) + (b^(2 m) E^((
      qL \[Rho])/\[Lambda]) m p^(-1 + m)
       RqL (E^((
         b^m (qH + 2 qL))/((b^m + p^m) \[Lambda])) (qH - 
           qL) (-1 + \[Alpha]) + 
        E^(b/\[Lambda]) (-E^(((2 b)/\[Lambda])) qL - 
           2 E^((b^
             m (qH + qL))/((b^m + p^m) \[Lambda])) (qH - 
              qL) (-1 + \[Alpha]) - 
           2 E^((b + (b^m qL)/(b^m + p^m))/\[Lambda])
             qL (-1 + \[Alpha]) + 
           E^((2 qL \[Rho])/\[Lambda]) qL (-1 + \[Alpha]) - 
           E^((2 b^m qH)/((b^m + p^m) \[Lambda])) qL \[Alpha] + 
           E^((b + (b^m qH)/(
             b^m + p^m))/\[Lambda]) (-qH + 
              qL + (qH + qL) \[Alpha]))))/((b^m + p^m)^3 (E^((
        2 b)/\[Lambda]) - E^((b + (b^m qH)/(b^m + p^m))/\[Lambda]) - 
        E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (-1 + \[Alpha]) + 
        E^((2 qL \[Rho])/\[Lambda]) (-1 + \[Alpha]))^2 \[Lambda]) < 
  0, 100 > qH > qL > b > p > 0 &&  0 < \[Rho] < 1 && 
  2 > \[Lambda] > 0 && 0 < \[Alpha] < 1 && 2 > m > 1]
POSTED BY: Li Siqi
6 Replies
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POSTED BY: Gianluca Gorni

Thanks, but there are some questions about your answers. Is your code presented a simulation about one parameter? I cannot understand the meaning of plot.

Could you please tell me detailedly? (I have read the official document, but I learned MMA from yesterday.)

And I will post some other region of the parameters. They are:

qH > qL > 0 && b > 0 && p > 0 && 0 < \[Rho] < 1 && \[Lambda] > 0 && 
 0 < \[Alpha] < 1 && m > 0

which should be put after FindInstance as the first part.

Thanks, bro!!
POSTED BY: Li Siqi
POSTED BY: Gianluca Gorni
POSTED BY: Li Siqi

Also the new expression has no fixed sign, it appears:

expr = D[qL b^
        m/(b^m + 
         p^m) Exp[(qL b^m/(b^m + p^m) - 
           b)/\[Lambda]]/(Exp[(qL b^m/(b^m + p^m) - 
             b)/\[Lambda]] + ((1 - 
              Exp[(qH b^m/(b^m + p^m) - b)/\[Lambda]]) (1 - 
              Exp[(qL b^m/(b^m + p^m) - 
                  b)/\[Lambda]]))/(1 - \[Alpha] Exp[(qH b^
                    m/(b^m + p^m) - 
                 b)/\[Lambda]] - (1 - \[Alpha]) Exp[(qL b^
                    m/(b^m + p^m) - b)/\[Lambda]])) + p, p] // 
   Simplify;
inst = FindInstance[
  100 > qH > qL > b > p > 0 && 0 < \[Rho] < 1 && 2 > \[Lambda] > 0 && 
   0 < \[Alpha] < 1 && 2 > m > 1 && Element[qH | qL | b | p, Integers],
  {qH, qL, b, p, \[Rho], \[Lambda], \[Alpha], m}, 2]
N[Simplify[expr /. inst], 5]
POSTED BY: Gianluca Gorni

Thanks, bro. I will reconsider my function form.

Thank you again for you generous help！！
POSTED BY: Li Siqi
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