Problem 1465

Assume, by the sake of contradiction, that there exist $b, c \in \mathbb{R}$, $b < c$, such that $\left|\frac{x+y}{x-y}\right| \le a$, for every $x, y \in (b, c)$, $x \ne y$, and consider $x \in (b, c)$, $x \ne 0$.

If $x > 0$, there exist infinitely many positive integers $n$ such that $n > \frac{1}{x-b}$, or equivalently, such that $x - \frac{1}{n} > b$. For each such number $n$, choose $y_n = x - \frac{1}{n} \in (b, c)$. Since $\left|\frac{x+y_n}{x-y_n}\right| \le a$, it follows that $x \le \frac{1+a}{2n}$, so $0 < n \le \frac{1+a}{2x}$ for infinitely many positive integers $n$, a contradiction.

If $x < 0$, there are infinitely many positive integers $n$ such that $n > \frac{1}{c-x}$, or equivalently, such that $x + \frac{1}{n} < c$. For each such number $n$, choose $z_n = x + \frac{1}{n} \in (b, c)$. Since $\left|\frac{x+z_n}{x-z_n}\right| \le a$, it follows that $-\frac{1+a}{2n} \le x < 0$, so $0 < n \le -\frac{1+a}{2x}$ for infinitely many positive integers $n$, again a contradiction, and the conclusion follows.