IMC 1999 · Day 2 · P10

Assume that such a function exists. The initial inequality can be written in the form $f(x) - f(x+y) \ge f(x) - \dfrac{f^2(x)}{f(x) + y} = \dfrac{f(x) y}{f(x) + y}$. Obviously, $f$ is a decreasing function. Fix $x > 0$ and choose $n \in \mathbb{N}$ such that $n f(x+1) \ge 1$. For $k = 0, 1, \dots, n-1$ we have $$f \left( x + \frac{k}{n} \right) - f \left( x + \frac{k+1}{n} \right) \ge \frac{f \left( x + \frac{k}{n} \right)} {n f \left( x + \frac{k}{n} \right) + 1} \ge \frac{1}{2n}.$$ The additon of these inequalities gives $f(x+1) \le f(x) - \frac{1}{2}$. From this it follows that $f(x + 2m) \le f(x) - m$ for all $m \in \mathbb{N}$. Taking $m \ge f(x)$, we get a contradiction with the conditon $f(x) > 0$.