# Is this expression positive or negative given the assumptions?

Consider:

Expre78 = ( b βa (μ (-1 + ν) - ξ) + μ (μ + ξ + ρ) (γa + μ + σ))/(μ (μ + ξ + ρ) σ);

Assuming[
{b > 0, βa > 0, μ > 0, 0 < ν < 1, βi > 0, ξ > 0, ρ > 0, γa > 0, σ > 0},
Reduce[$Assumptions \[Implies] Expre78 >= 0, {}, Reals] ]  Is this expression positive or negative given the assumptions? Edit: if r0<1 then i want to check whether Expre78 <0 r0=(b (\[Mu] - \[Mu] \[Nu] + \[Xi]) (\[Beta]a (\[Alpha] + \[Gamma]i + \ \[Eta] + \[Mu]) + \[Beta]i \[Sigma]))/(\[Mu] (\[Alpha] + \[Gamma]i + \ \[Eta] + \[Mu]) (\[Mu] + \[Xi] + \[Rho]) (\[Gamma]a + \[Mu] + \ \[Sigma])); Expre78 = (b \[Beta]a (\[Mu] (-1 + \[Nu]) - \[Xi]) + \[Mu] (\[Mu] + \ \[Xi] + \[Rho]) (\[Gamma]a + \[Mu] + \[Sigma]))/(\[Mu] (\[Mu] + \[Xi] \ + \[Rho]) \[Sigma]); Assuming[{b > 0, \[Beta]a > 0, \[Beta]i > 0, \[Mu] > 0, 0 < \[Nu] < 1, \[Gamma]a > 0, \[Gamma]i > 0, \[Xi] > 0, \[Rho] > 0, \[Eta] > 0, \[Alpha] > 0, \[Sigma] > 0, r0 < 1}, Reduce[$Assumptions \[Implies] Expre78 <= 0, {}, Reals]]

• Did you try FindInstance[{Expre78 < 0 && b > 0 && \[Beta]a > 0 && \[Mu] > 0 && 0 < \[Nu] < 1 && \[Xi] > 0, \[Rho] > 0, \[Gamma]a > 0, \[Sigma] > 0}, {b, \[Beta]a, \[Mu], \[Nu], \[Xi], \[Rho], \[Gamma]a, \ \[Sigma]}]? Commented Jul 12, 2022 at 14:21
• @user293787 I amended the assumptions
– Math
Commented Jul 12, 2022 at 15:37

Clear["Global*"]

Expre78 = (b βa (μ (-1 + ν) - ξ) + μ (μ + ξ + ρ) (γa + μ + σ))/
(μ (μ + ξ + ρ) σ);


It can be either.

To be nonnegative

Assuming[{b > 0, βa > 0, μ > 0,
0 < ν <= 1, ξ > 0, ρ > 0, γa > 0, σ > 0},
Reduce[Expre78 >= 0] // Simplify]

(* b βa (μ - μ ν + ξ) <= μ (μ + ξ + ρ) (\
γa + μ + σ) *)

vars = Variables[Level[Expre78, {-1}]];


Nonnegative values

Expre78 >= 0 /.
FindInstance[{Expre78 >= 0, b > 0, βa > 0, μ > 0,
0 < ν <= 1, ξ > 0, ρ > 0, γa > 0, σ > 0}, vars, 5]

(* {True, True, True, True, True} *)


Negative values

Expre78 < 0 /.
FindInstance[{Expre78 < 0, b > 0, βa > 0, μ > 0,
0 < ν <= 1, ξ > 0, ρ > 0, γa > 0, σ > 0}, vars, 5]

(* {True, True, True, True, True} *)


EDIT: For the revised question

Clear["Global*"]

r0 = (b (μ - μ ν + ξ) (βa (α + γi + η + μ) + βi σ))/
(μ (α + γi + η + μ) (μ + ξ + ρ) (γa + μ + σ));

Expre78 = (b βa (μ (-1 + ν) - ξ) + μ (μ + ξ + ρ) *
(γa + μ + σ))/(μ (μ + ξ + ρ) σ);

vars = Variables[Level[Expre78 && r0, {-1}]]

{b, α, βa, βi, γa, γi, η, μ, ν, ξ, ρ, σ}


There are no nonpositive cases

FindInstance[{Expre78 <= 0, b > 0, βa > 0, βi > 0, μ > 0,
0 < ν < 1, γa > 0, γi > 0, ξ > 0, ρ > 0, η >
0, α > 0, σ > 0, r0 < 1}, vars]

(* {} *)


or

Reduce[{Expre78 <= 0, b > 0, βa > 0, βi > 0, μ > 0,
0 < ν < 1, γa > 0, γi > 0, ξ > 0, ρ > 0, η >
0, α > 0, σ > 0, r0 < 1}]

(* False *)


There is at least one positive case

inst = FindInstance[{Expre78 > 0,
b > 0, βa > 0, βi > 0, μ > 0,
0 < ν < 1, γa > 0, γi > 0, ξ > 0, ρ > 0,
η > 0, α > 0, σ > 0, r0 < 1}, vars]

(* {{b -> 7/2, α -> 1, βa -> 1, βi -> 3/2, γa -> 1,
γi -> 1, η -> 1, μ -> 1, ν -> 1/2, ξ -> 1,
ρ -> 1, σ -> 1}} *)

Expre78 /. inst[[1]]

5/4

• Thank you. Could you attach examples showing where it is positive and negative?
– Math
Commented Jul 12, 2022 at 14:25
• @Math please make an effort to understand the answer by BobHanlon, the examples are created there by using FindInstance. Commented Jul 12, 2022 at 14:26
• Look at the FindInstance expressions shown prior to evaluating the ReplaceAll Commented Jul 12, 2022 at 14:27
• @BobHanlon, I have added an addendum since I missed on important assumption.
– Math
Commented Jul 12, 2022 at 15:19
• @BobHanlon Thank you for the update. I have one question; if Reduce[{Expre78 > 0, b > 0, βa > 0, βi > 0, μ > 0, 0 < ν < 1, γa > 0, γi > 0, ξ > 0, ρ > 0, η > 0, α > 0, σ > 0, r0 < 1}] outputs false then why when I change the reduce statement to Expre78 > 0 (or even Expre78 >= 0) it shows conditions? Shouldn't it output true?
– Math
Commented Jul 13, 2022 at 10:55