Relating an expression to two similar ones

Question

Is it possible to express $C$ solely in terms of $A$ and $B$, where

$$A = \dfrac{m}{x+z}, B = \dfrac{n}{y+z}, C=\dfrac{m+n}{x+y+z}$$

and $m,n,x,y,z>0 \ ?$

If not, how close can I get?

Answer 1

Here is a proof by counterexample that you cannot express $C$ solely in terms of $A$ and $B$:

Let's take $ a = \frac{1}{6}$ and $B = \frac{1}{8}$ and get these values in two ways: $$ m = n = 1 \text{, } (x,y,z) = (1,5,3) \\ $$ and $$ m' = 1 \text{, } n' = \frac{8}{7} \text{, } (x',y',z') = (2,4,3) $$ Then using $m,n,x,y,z$ we get $C = \frac{2}{9}$.

And using $m',n',x',y',z'$ we get $C' = \frac{5}{21}$. There are two different values of $C$ for identical values of $A$ and $B$.

Let's phrase your second question more precisely:

Let
$$ C_\max(A,B) \equiv \max_{m,n,x,y,z} \frac{m+n}{x+y+z} : \left( \frac{m}{x+z}=A \right) \wedge $\left( \frac{n}{y+z}=B \right) $$ and $$ C_\max(A,B) \equiv \min_{m,n,x,y,z} \frac{m+n}{x+y+z} : \left( \frac{m}{x+z}=A \right) \wedge $\left( \frac{n}{y+z}=B \right) $$ Find $C_\max(A,B) - C_\min(A,B) $ as a function of $A$ and $B$.

Unfortunately, you can't get close at all to the value of $C$ in terms of $A$ and $B$ because for fixed $A$ and $B$ (unless $A = B = 0$), $$ C = A + (B-A) \frac{y}{x+y+z} + B \frac{z}{x+y+z} $$ so choosing $$ x = -1-(k-1)\epsilon \\ y = \frac{1}{2} + \epsilon \\ z = \frac{1}{2} + k \epsilon $$ and selecting $m,n$ to give the correct values of $A,B$,both $(B-A) \frac{y}{x+y+z}$ and $B \frac{z}{x+y+z}$ blow up as $\epsilon$ goes to zero, if the corresponding numerator is non-zero. For any particular values of $(A,B)$ other than $(0,0)$, we can choose $k$ such that those two terms are not negatives of one another. So we can make $$C$$ arbitrarily large or arbitrarily small, for any values of $A$ and $B$.