Saudi Booklet

First, choose $x = \frac{n+1}{n+2}$, $y = \frac{n}{n+1}$ with $n \in \mathbb{Z}^+$ then $xy = \frac{n}{n+2} < 1$. Substituting into the given inequality, we have
$$\begin{cases} \frac{n+1}{n+2} + \frac{n}{n+1} < c + \frac{n}{n+2} \\ \iff c > \frac{n}{n+1} + \frac{1}{n+2}. \end{cases}$$
Here, for $n \to +\infty$ we have $c \ge 1$. Next, we will prove the inequality with $c = 1$: $\{x\} + \{y\} < 1 + xy$ for all $x, y > 0$ and $xy < 1$. Indeed, we have
$$(1 - \{x\})(1 - \{y\}) > 0 \implies \{x\} + \{y\} < 1 + \{x\}\{y\} \le 1 + xy.$$
This inequality is true. Therefore, the minimum value we need to find is $c = 1$.