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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
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Thanks, bro. I will reconsider my function form.

Thank you again for you generous help！！
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

Sorry, sir, I think I posted the wrong code about my question. And I posted the complete code about my question:

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]

And I want to judge the sign of the outcome above. Thank you for your carefulness, otherwise I will never find my wrong.

Please help me judge the sign of the outcome above. Thaaaaaaaanks!!
POSTED BY: Li Siqi

I tried to make your expression more manageable by setting most variable to constants. Your expression is linear in the variable RqL, and you make no assumption on RqL. FindInstance finds some constant values for the other parameters that satisfy your constraints. I replace those values into the expression and plot the result as a function of RqL. The plot changes sign. This is an indication that your expression has no definite sign. Plot makes numerical approximations, so that this is no mathematical proof, but it points to a direction for further investigation.
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

It does not seem that your expression has a definite sign:

inst = FindInstance[
   100 > qH > qL > b > p > 0 && 0 < \[Rho] < 1 && 2 > \[Lambda] > 0 &&
     0 < \[Alpha] < 1 && 2 > m > 1, {m, qH, qL, b, 
    p, \[Rho], \[Lambda], \[Alpha]}, Reals, 3];
expr = 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]);
Plot[expr /. inst[[2]], {RqL, -1, 1}]
POSTED BY: Gianluca Gorni
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