IMC 2001 · Day 2 · P11

Suppose that there exists a function satisfying the inequality. If $f(f(x)) \le 0$ for all $x$, then $f$ is a decreasing function in view of the inequalities $f(x + y) \ge f(x) + y f(f(x)) \ge f(x)$ for any $y \le 0$. Since $f(0) > 0 \ge f(f(x))$, it implies $f(x) > 0$ for all $x$, which is a contradiction. Hence there is a $z$ such that $f(f(z)) > 0$. Then the inequality $f(z + x) \ge f(z) + x f(f(z))$ shows that $\lim\limits_{x \to \infty} f(x) = +\infty$ and therefore $\lim\limits_{x \to \infty} f(f(x)) = +\infty$. In particular, there exist $x, y > 0$ such that $f(x) \ge 0$, $f(f(x)) > 1$, $y \ge \dfrac{x + 1}{f(f(x)) - 1}$ and $f(f(x + y + 1)) \ge 0$. Then $f(x + y) \ge f(x) + y f(f(x)) \ge x + y + 1$ and hence $$\begin{align*} f(f(x+y)) &\ge f(x+y+1) + \bigl( f(x+y) - (x+y+1) \bigr) f(f(x+y+1)) \ge \\ &\ge f(x+y+1) \ge f(x+y) + f(f(x+y)) \ge \\ &\ge f(x) + y f(f(x)) + f(f(x+y)) > f(f(x+y)). \end{align*}$$ This contradiction completes the solution of the problem.