Increasing the sum of two ratios by shifting material between the denominators Let $x_1,x_2 > 0$ and let $y_1,y_2 > \epsilon > 0$. Let $R$ be the sum of ratios:
$$
R = \frac{x_1}{y_1} + \frac{x_2}{y_2}
$$
Now, steal $\epsilon$ from one of the denominators and donate it to the other one. This can be done in two ways, so let:
$$
R^\prime = \frac{x_1}{y_1-\epsilon} + \frac{x_2}{y_2+\epsilon}\\
R^{\prime\prime} = \frac{x_1}{y_1+\epsilon} + \frac{x_2}{y_2-\epsilon}
$$
By looking at many examples with randomly-generated numbers, it seems to be the case that at least one of $R^\prime$ and $R^{\prime\prime}$ is larger than $R$; sometimes both.
I would like to prove this result, and I would like to find a simple condition that tells me which one(s) of $R^\prime$ and $R^{\prime\prime}$ will be larger than $R$. I have tried for a while but I cannot get anything nice to fall out of the messy algebra.
 A: After playing around with it on more sleep and more caffeine, I managed to get it to work. My algebra had to get pretty messy before a nice solution would fall out, so I'm putting it here in its full goriness. I would be interested to see if anyone can get the same result with simpler algebra.
We have $R^\prime > R$ if:
\begin{align}
\frac{x_1}{y_1-\epsilon} + \frac{x_2}{y_2+\epsilon} &>  \frac{x_1}{y_1} + \frac{x_2}{y_2}\\
\frac{(x_1 y_2 + x_2 y_1)+(x_1 - x_2)\epsilon}{y_1y_2+(y_1-y_2)\epsilon +\epsilon^2} &> \frac{x_1 y_2 + x_2 y_1}{y_1 y_2}\\
\frac{(x_1 y_2 + x_2 y_1)+(x_1 - x_2)\epsilon}{x_1 y_2 + x_2 y_1} &> \frac{y_1y_2+(y_1-y_2)\epsilon +\epsilon^2}{y_1 y_2}\\
1 + \frac{x_1-x_2}{x_1 y_2 + x_2 y_1}\epsilon &> 1 + \frac{y_1 - y _2}{y_1 y_2}\epsilon + \frac{1}{y_1 y_2}\epsilon^2\\
\frac{x_1 - x_2}{x_1 y_2 + x_2 y_1} &> \frac{y_1 - y_2 + \epsilon}{y_1 y_2}\\
(x_1 - x_2)y_1 y_2 &> (x_1 y_2 + x_2 y_1)(y_1 - y_2 - \epsilon)\\
x_1 y_1 y_2 - x_2 y_1 y_2 &> x_1 y_1 y_2 - x_1 y_2 (y_2 + \epsilon) + x_2 y_1 (y_1 - \epsilon) - x_2 y_1 y_2\\
x_1 y_2 (y_2 + \epsilon) &> x_2 y_1 (y_1 - \epsilon)\\
\frac{x_1/y_1}{x_2/y_2} &> \frac{y_1 - \epsilon}{y_2 + \epsilon}
\end{align}
By symmetry, we have $R^{\prime\prime}>R$ if we substitute $x_1 \leftrightarrow x_2 $ and $y_1 \leftrightarrow y_2 $ in the above result, which gives us:
\begin{align}
\frac{x_2/y_2}{x_1/y_1} &> \frac{y_2 - \epsilon}{y_1 + \epsilon}\\
\frac{x_1/y_1}{x_2/y_2} &< \frac{y_1 + \epsilon}{y_2 - \epsilon}
\end{align}
Note that we always have $\frac{y_1 - \epsilon}{y_2 + \epsilon} < \frac{y_1 + \epsilon}{y_2 - \epsilon}$ (numerator is smaller and denominator is larger), which gives us the following, full picture:

A: Note that $R = \frac{x_1y_2 + x_2y_1}{y_1y_2}$. Using $y_2 \leftarrow y_2 + \epsilon$ and $y_1 \leftarrow y_1 - \epsilon$ yields $$R^{'} = \frac{x_1y_2 + x_2 y_1 + (x_1 - x_2)\epsilon}{y_1y_2 + (y_1 - y_2)\epsilon - \epsilon^2}.$$ Therefore, if $x_1 \geq x_2$ and $y_1 \leq y_2$, we have that $R^{'} \geq R$ because the numerator is greater than before and the denominator is smaller than before. 
Can you work the other cases out?
