Problem 1513

Therefore the maximum value of the sum of the fractions is attained when $y$ is minimum, which is $1$. When $y = 1$, since $ac < xy$, the optimum value for $x$ is $ac + 1$ (we want to choose $x$ so that it is as small as possible). Thus for fixed $a, c$, the maximum value is
$$\frac{a}{a+ac+1} + \frac{c}{c+1} = 1 - \frac{1}{(a+ac+1)(c+1)}.$$
By checking all pairs $(a, c) = (1, 19), (2, 18), \dots, (10, 10)$, we find that the maximum value of $(a + ac + 1)(c + 1)$ is $1386$, attained when $(a, c) = (7, 13)$. Therefore the maximum value is $1 - \frac{1}{1386} = \frac{1385}{1386}$.

We shall use the following result. Let $p, q, k$ be positive integers. Then $\frac{p}{p+k} < \frac{q}{q+k}$ iff $p < q$.

First fix $a, c$. Without loss of generality, we assume that $a \le c$. Let $b = a + x$ and $d = c + y$ where $x, y$ are positive integers. Then it is easy to see that
$$\frac{a}{a+x} + \frac{c}{c+y} \ge \frac{a}{a+y} + \frac{c}{c+x}$$
if $y \le x$. Thus we may assume that $y \le x$. Also
$$1 > \frac{a}{b} + \frac{c}{d} = \frac{a}{a+x} + \frac{c}{c+y} = \frac{ac + ay + ac + cx}{ac + ay + xy + cx}$$
Thus $ac < xy$ and $xy - ac$ is the difference of the denominator and numerator of the fraction on the RHS. Thus if $x', y'$ are positive integers such that $y' \le x'$ and $xy = x'y'$, then $|y - y'| \le |x - x'|$. Therefore
$$$$
$$\begin{align*} \frac{a}{a+x} + \frac{c}{c+y} &\le \frac{a}{a+x'} + \frac{c}{c+y'} \\ \Leftrightarrow (a+x)(c+y) &\le (a+x')(c+y') \\ \Leftrightarrow a(y - y') &\le c(x' - x) \\ \Leftrightarrow y &\ge y \quad \text{and} \quad x' \ge x \end{align*}$$