Assume that the numbers \( b \) and \( c \) exist as described in the statement. Let \( x \in (b, c) \), \( x \neq 0 \). We have two cases: I. \( x > 0.\) There exist infinitely many natural numbers \( n \) such that \( n > \frac{1}{x - b} \), so that \( x - \frac{1}{n} > b \). For each such number \( n \), we choose \( y_n = x - \frac{1}{n} \in (b, c) \). The hypothesis relationship, applied for each pair \( (x, y_n) \), leads to \[x \leq \frac{1 + a}{2n}.\]We obtain \[0 < n \leq \frac{1 + a}{2x}\]for infinitely many natural numbers \( n \), which is false. II. \( x < 0.\) There exist infinitely many natural numbers \( n \) such that \( n > \frac{1}{c - x} \), so that \( x + \frac{1}{n} < c \). For each such number \( n \), we choose \( z_n = x + \frac{1}{n} \in (b, c) \). The hypothesis relationship, applied for each pair \( (x, z_n) \), leads to \[ -\frac{1 + a}{2n} \leq x < 0. \]We obtain \[ 0 < n \leq -\frac{1 + a}{2x} \]for infinitely many natural numbers \( n \), which is false. From I and II, the conclusion follows.