IMO 1J, Mock Exam 1

All functions $f(x) = \frac{c}{x}$ for some $c > 0$.

Let $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^k(x) = f(f(\cdots f(x)\cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^0(x) = x$. Substituting $y = x$ gives
$$x \geq f^2(x).$$
Substituting $x = f(y)$ instead leads to $f(y) + f^2(y) \geq y + f^3(y)$, or equivalently
$$f(y) - f^3(y) \geq y - f^2(y).$$
We can generalize this inequality. If we replace $y$ by $f^{n-1}(y)$ in the above inequality, we get
$$f^n(y) - f^{n+2}(y) \geq f^{n-1}(y) - f^{n+1}(y),$$
for every $y \in \mathbb{R}_{>0}$ and for every integer $n \geq 1$. In particular, $f^n(y) - f^{n+2}(y) \geq y - f^2(y) \geq 0$ for every $n \geq 1$. Hereafter consider even integers $n = 2m$. Observe that
$$y - f^{2m}(y) = \sum_{i=0}^{m-1} f^{2i}(y) - f^{2i+2}(y) \geq m(y - f^2(y)).$$
Since $f$ takes positive values, it holds that $y - f^{2m}(y) < y$ for every $m \geq 1$. So, we have proved that $y > m(y - f^2(y))$ for every $y \in \mathbb{R}_{>0}$ and every $m \geq 1$. Since $y - f^2(y) \geq 0$, this holds if and only if
$$f^2(y) = y$$
for every $y \in \mathbb{R}_{>0}$. The original inequality becomes
$$x f(x) \geq y f(y)$$
for every $x, y \in \mathbb{R}_{>0}$. Hence, $x f(x)$ is a constant. We conclude that $f(x) = \frac{c}{x}$ for some $c > 0$.

Let $f : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. As in Solution 1, we prove that
$$f^n(y) \ge f^{n+2}(y)$$
for every $y \in \mathbb{R}_{>0}$ and every $n \ge 0$. Since $f$ takes positive values, this implies that
$$y f(y) \ge f(y) f^2(y) \ge f^2(y) f^3(y) \ge \dots$$
In other words, $y f(y) \ge f^n(y) f^{n+1}(y)$ for every $y \in \mathbb{R}_{>0}$ and every $n \ge 1$.
We replace $x$ by $f^n(x)$ in the original inequality and get
$$f^n(x) - f^{n+2}(x) \ge \frac{y f(y) - f^n(x) f^{n+1}(x)}{f(y)}.$$
Using that $x f(x) \ge f^n(x) f^{n+1}(x)$, we obtain
$$f^n(x) - f^{n+2}(x) \ge \frac{y f(y) - x f(x)}{f(y)}$$
for every $n \ge 0$. The same trick as in Solution 1 gives
$$x > x - f^{2m}(x) = \sum_{i=0}^{m-1} f^{2i}(x) - f^{2i+2}(x) \ge m \cdot \frac{y f(y) - x f(x)}{f(y)}$$
for every $x, y \in \mathbb{R}_{>0}$ and every $m \ge 1$. Possibly permuting $x$ and $y$, we may assume that $y f(y) - x f(x) \ge 0$ then the above inequality implies $x, f(x) = y f(y)$. We conclude as in Solution 1. $\square$