# Need help with solving inequality under assumptions [duplicate]

I seriously need help for my research. I need to solve an inequality with several assumptions. I tried many times with Reduce but it did not work out properly. Sometimes the results I get are simply wrong or I encounter inconsistencies like Mathematica telling me that the result is neither positive nor equal to zero but only negative under certain assumptions. (though I assumed that all variables are in the Reals). It is of high importance that I use Mathematica correctly, because if I can't trust my results I can't publish them. What I need to do is the following: I need to know if either:

-(2*F*T-2*r*F-m)/((2*r-2)FT+(2*c-2*m)*r+2*m-2*c)-(((2*r-2)*F+m)/((2*r-2)*F+(2*c-2*m)*r+2*m-2*c))>0

or

-(2*F*T-2*r*F-m)/((2*r-2)FT+(2*c-2*m)*r+2*m-2*c)-(((2*r-2)*F+m)/((2*r-2)*F+(2*c-2*m)*r+2*m-2*c))<0

or

-(2*F*T-2*r*F-m)/((2*r-2)FT+(2*c-2*m)*r+2*m-2*c)-(((2*r-2)*F+m)/((2*r-2)*F+(2*c-2*m)*r+2*m-2*c))==0

under the following assumptions:

0 < a <= 1/2 && m > 0 && 0 <= T < 1 && 0 <= d <= 1 && F > 0 &&
0 < r < 1 && c >= 0 && a [Element] Reals && T [Element] Reals && d [Element] Reals && F [Element] Reals && r [Element] Reals && m [Element] Reals && c [Element] Reals && m > c && -2 c + m - 2 F r > 0

or in other terms if -(2*F*T-2*r*F-m)/((2*r-2)FT+(2*c-2*m)*r+2*m-2*c) is bigger, smaller or equal to ((2*r-2)*F+m)/((2*r-2)*F+(2*c-2*m)*r+2*m-2*c) or under which additional assumptions this is the case. (Above I subtracted the second from the first)

Until now I always tried to work with the following:

LogicalExpand@
FullSimplify[$Assumptions = 0 < a <= 1/2 && m > 0 && 0 <= T < 1 && 0 <= d <= 1 && F > 0 && 0 < r < 1 && c >= 0 && a ∈ Reals && T ∈ Reals && d ∈ Reals && F ∈ Reals && r ∈ Reals && m ∈ Reals && c ∈ Reals && m > c && -2 c + m - 2 F r >= 0; Reduce[-(2*F*T - 2*r*F - m)/((2*r - 2)*F*T + (2*c - 2*m)*r + 2*m - 2*c) - (((2*r - 2)*F + m)/((2*r - 2)*F + (2*c - 2*m)*r + 2*m - 2*c)) > 0 &&$Assumptions]]
(* (c + F < m && 2 (c + F r) < m) || (c + F > m && m < c + F T &&
2 (c + F r) < m) || 2 (c + F) <= m *)

LogicalExpand@
FullSimplify[$Assumptions = 0 < a <= 1/2 && m > 0 && 0 <= T < 1 && 0 <= d <= 1 && F > 0 && 0 < r < 1 && c >= 0 && a ∈ Reals && T ∈ Reals && d ∈ Reals && F ∈ Reals && r ∈ Reals && m ∈ Reals && c ∈ Reals && m > c && -2 c + m - 2 F r >= 0; Reduce[-(2*F*T - 2*r*F - m)/((2*r - 2)*F*T + (2*c - 2*m)*r + 2*m - 2*c) - (((2*r - 2)*F + m)/((2*r - 2)*F + (2*c - 2*m)*r + 2*m - 2*c)) < 0 &&$Assumptions]]
(* c + F > m && 2 (c + F r) < m && c + F T < m *)

LogicalExpand@
FullSimplify[$Assumptions = 0 < a <= 1/2 && m > 0 && 0 <= T < 1 && 0 <= d <= 1 && F > 0 && 0 < r < 1 && c >= 0 && a ∈ Reals && T ∈ Reals && d ∈ Reals && F ∈ Reals && r ∈ Reals && m ∈ Reals && c ∈ Reals && m > c && -2 c + m - 2 F r >= 0; Reduce[-(2*F*T - 2*r*F - m)/((2*r - 2)*F*T + (2*c - 2*m)*r + 2*m - 2*c) - (((2*r - 2)*F + m)/((2*r - 2)*F + (2*c - 2*m)*r + 2*m - 2*c)) == 0 &&$Assumptions]]
(* (2 (c + F r) == m && 2 r > 1) || (2 r == 1 &&
2 (c + F r) == m && F < m) || (2 (c + F r) == m &&
m + 2 F r > 2 F && 2 r < 1) || (2 (c + F r) == m && 2 r < 1 &&
m + 2 F r < 2 F && 2 F T != m + 2 F r) *)


Though, trough other trials of solving with other programs and by the pure logic of my model I believe to know that these results are wrong under my assumptions. Mathematica says that the term can also be negative under certain assumptions. But I know in my model and under this assumptions it should be impossible. I also found a clearly contradicting result when solving by hand. And sometimes Mathematica is even contridicting itself when I try to get deeper into it. I believe that Reduce does not use my assumptions properly. Now in this case I believe to know that Mathematica is wrong but I will have to do similar calculations with other terms in the future where I cannot predict results anymore, so it is very important for me to have a code that works properly also for the later phase of my calculations. So I need a general way to solve this problem properly not only for this case. I really hope someone can help me out here. Thanks beforehand.