1
$\begingroup$

I've got an inequality of 8 non-negative parameters (arising from a 5x5 matrix) and would like to find a condition under which my inequality holds. The condition would be an inequality too of these parameters. I'm not looking for the numerical values of the parameters, but trying to get an if and only if relation in a proof with a very large inequality to solve. For instance, given that all parameters are non-negative and 0<q<1, what condition should the parameters satisfy in order to have the following inequality:

      n*k*(β1*(1 - q) + β2*q)) - ((k + μ)*(γ1 + μ)*(γ2 + μ) - 
     β2*n*k*q*(γ1 + μ) - β1*n*k*(1 - q)*(γ2 + μ)) > 0)

Does anyone know if Mathematica can do this? Thank you.

$\endgroup$
2
$\begingroup$

First, don't use N as symbol, it's predfined in Mathematica

Try Reduce to solve the problem

expr = (n*
 k*(\[Beta]1*(1 - q) + \[Beta]2*
    q)) - ((k + \[Mu])*(\[Gamma]1 + \[Mu])*(\[Gamma]2 + \[Mu]) -\[Beta]2*n*k*q*(\[Gamma]1 + \[Mu]) - \[Beta]1*n*k*(1 - q)*(\[Gamma]2 + \[Mu])) ;
var = Variables[expr];
Reduce[Join[{expr > 0}, Map[# > 0 &, var]]] // Simplify


(*k > -((\[Mu] (\[Gamma]1 + \[Mu]) (\[Gamma]2 +\[Mu]))/(-n q \[Beta]2 \(1 + \[Gamma]1 + \[Mu]) + (\[Gamma]1 + \[Mu])(\[Gamma]2 + \[Mu]) + n (-1 + q) \[Beta]1 (1 + \[Gamma]2 +\[Mu]))) 
&&n > ((\[Gamma]1 + \[Mu]) (\[Gamma]2 + \[Mu]))/(q \[Beta]2 (1 + \[Gamma]1 + \[Mu]) - (-1 +q) \[Beta]1 (1 + \[Gamma]2 + \[Mu])) 
&& \[Beta]2 >0 && \[Gamma]1 > 0 && \[Gamma]2 > 0 && \[Mu]> 0 
&& ((\[Beta]1 > (\[Beta]2 (1 + \[Gamma]1 + \[Mu]))/(1 + \[Gamma]2 + \[Mu]) 
&&0 < q < (\[Beta]1 (1 + \[Gamma]2 + \[Mu]))/(-\[Beta]2(1 + \[Gamma]1 + \[Mu]) + \[Beta]1 (1 + \[Gamma]2 + \[Mu]))) || (q > 0 && 0 < \[Beta]1 <= (\[Beta]2 (1 + \[Gamma]1 + \[Mu]))/(1 + \[Gamma]2 + \[Mu])))*)
| improve this answer | |
$\endgroup$
  • $\begingroup$ It isn't actually an Ν, it's a capital Nu ($\nu$), but it's confusing either way and might lead to somebody typing a real N. $\endgroup$ – flinty Jul 9 at 11:52
  • $\begingroup$ Please edit your question and change N to Nu! $\endgroup$ – Ulrich Neumann Jul 9 at 14:28
  • $\begingroup$ It's not my question - @Vicky please edit it. $\endgroup$ – flinty Jul 9 at 14:41
  • $\begingroup$ @flinty Sorry, but how do you know that N reads as Nu??? $\endgroup$ – Ulrich Neumann Jul 9 at 14:42
  • $\begingroup$ Copy an paste OP's code Ν*k. See that the first Ν is blue and has serifs? Copy and paste this into Mathematica and you'll see the difference: Ν*N. $\endgroup$ – flinty Jul 9 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.