I have the following equations and inequalities:
$1 = A' + B'$
$1 = A + B + C$
$A \le A'$
$B \le B'$
All variables are bounded below by zero and above by one.
I wonder if I can find an analytic expression for the upper and lower bounds for the difference $A'(1 - C) - A$. The Monte-Carlo experiment shows that it can be either positive or negative:
and it seems like there are some nice looking bounding curves.
First note $A'(1-C)-A=A'(A+B)-A$.
For the upper bound clearly we need to take $A=0$, $B=B'$ and $C=A'$. In which case the problem comes down to finding the maximum value of $A'B'=A'-A'^{2}$ which is obtained if $A'=\frac{1}{2}$ so an upper bound is $\frac{1}{4}$. For a lower bound clearly we need to take $A=A'$, $B=0$ and $C=B'$, so we have to find the minimum value of $A^{2}-A$ for $0\leq A\leq 1$, which is obtained if $A=\frac{1}{2}$. So a lower bound is $-\frac{1}{4}$.
So $-\frac{1}{4}\leq A'(1-C)-A\leq\frac{1}{4}$.