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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:

Image

and it seems like there are some nice looking bounding curves.

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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 $-\frac{1}{4}\leq A'(1-C)-A\leq\frac{1}{4}$.

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