I have the following equations and inequalities: 1 = A ′ + B ′ 1...

Step 1
First note $A^{\prime }(1-C)-A=A^{\prime }(A+B)-A$
For the upper bound clearly we need to take $A=0$ , $B=B^{\prime }$ and $C=A^{\prime }$ . In which case the problem comes down to finding the maximum value of $A^{\prime }{B}^{\prime }=A^{\prime }-A^{\prime 2}$ which is obtained if $A^{\prime }=\frac{1}{2}$ so an upper bound is $\frac{1}{4}$. For a lower bound clearly we need to take $A=A^{\prime }$ , $B=0$ and $C=B^{\prime }$ , so we have to find the minimum value of $A^2-A$ for $0\le A\le 1$ , which is obtained if $A=\frac{1}{2}$. So a lower bound is $-\frac{1}{4}$ .
So $-\frac{1}{4}\le A^{\prime }(1-C)-A\le \frac{1}{4}$ .