I don't understand what the correct form of this inequality is

[handicapped_runner]

I have the following inequality:
[MATH] (C+Bk)(1-m)+B(m+f)(1-a)>0 [/MATH]I follow the following steps:
[MATH] (C+Bk)(1-m)>-B(m+f)(1-a)\\ C(1-m)>-B(m+f)(1-a)-Bk(1-mm)\\ \frac{C}{B}(1-m)>-(m+f)(1-a)-k(1-mm)\\ \frac{C}{B}>\frac{-(m+f)(1-a)-k(1-m)}{1-m}\\ \frac{-C}{B}<\frac{(m+f)(1-a)+k(1-m)}{(1-m)}\\ [/MATH]Is this correct? But what if I do the opposite, that is:
[MATH] B(m+f)(1-a)>-(C+Bk)(1-m)\\ B(m+f)(1-a)>-C(1-m)-Bk(1-m)\\ B((m+f)(1-a)+k(1-m))>-C(1-m)\\ B((m+f)(1-a)+k(1-m))>-C(1-m)\\ B>-C\frac{(1-m)}{((m+f)(1-a)+k(1-m))}\\ \frac{B}{-C}<\frac{(1-m)}{((m+f)(1-a)+k(1-m))}\\ [/MATH]They are clearly describing different things so I feel like I'm doing something wrong, but I cannot figure out what and which result is the correct one. Sorry, not a mathematician, I hope this isn't a very stupid question. C and B are positive(>0), while the other variables are between 0 and 1.

 

[Dr.Peterson]

If everything is positive, and a and m are between 0 and 1, then in each case the left-hand side is negative and the right-hand side is positive, so both inequalities are necessarily true (and not very informative). Without looking more closely, it appears that B and C are not really restricted by the inequality.

But all your work looks good!

 

[AmandasMathHelp]

I know the end results look different and at first glance I thought about the rule that when you flip a fractional inequality you have the change the sign. (If a/b <c/d then b/a > d/c) but that rule doesn't apply the same when a negative is involved.

Looking at them, on both their left sides they would have a negative number and both the right sides would be a positive number so clearly the left sides being less than the right are all good for both of them. Nice work!