# 2022 IMO Problems/Problem 2

## Problem

Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying

$$xf (y) + yf (x) \le 2$$.

## Solution

https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]

https://youtu.be/b5OZ62vkF9Y [Video Solution by little fermat]

Answer: The unique solution is the function $$f(x) = \frac{1}{x}$$ for every $$x \in \mathbb{R}^+$$. This function clearly satisfies the required property since the expression $$xf(y) + yf(x) = \frac{x}{y} + \frac{y}{x}$$ is greater than 2 for every $$y \neq x$$ (directly from AM-GM) and equal to 2 (with equality) for the unique value $$y = x$$.

Proof: Let's consider a solution based on some ideas we encountered in the preparation classes for the Olympiad, specifically involving auxiliary sets and functions with specific properties.

The fact that for every $$x \in \mathbb{R}^+$$, there exists a unique $$y \in \mathbb{R}^+$$ that satisfies the equation $$xf(y) + yf(x) \leq 2$$ can be equivalently expressed as follows: there exists a well-defined function $$g: \mathbb{R}^+ \to \mathbb{R}^+$$ given by $$g(x) := y$$, where $$y$$ is the one mentioned above. The well-definedness of this function is evident due to the existence and uniqueness, and it satisfies the equation $$P(x): \quad xf(g(x)) + f(x)g(x) \leq 2$$ while applying the same property for $$x \mapsto g(x)$$ gives another unique $$y := g(g(x))$$ such that $$g(x)f(y) + yf(g(x)) \leq 2$$. Therefore, we have $$xf(y) + yf(x) > 2$$ for all $$y \neq g(x)$$.

Since this inequality holds for $$y = x$$ (from $$xf(y) + yf(x) > 2$$), the uniqueness assumption implies that $$g(g(x)) = x$$, making $$g$$ an involution (hence bijective).

Generally, working with an involution naturally leads us to consider its fixed points, especially since we aim to show that $$g(x) = x$$ identically (which holds for the solution $$f(x) = \frac{1}{x}$$). Let's define the set of fixed points of $$g$$ as $$\mathcal{S} := \{ x \in \mathbb{R}^+ \mid g(x) = x \}$$ and show that $$\mathcal{S} = \mathbb{R}^+$$ is the entire domain.

Assume for a contradiction that some $$x \notin \mathcal{S}$$ is not a fixed point, i.e., $$x \neq g(x)$$. Then, the inequality $$2xf(x) > 2$$ (derived from $$y \mapsto x$$) holds, implying $$f(x) > \frac{1}{x}$$. Similarly, $$x \notin \mathcal{S}$$ implies $$g(x) \notin \mathcal{S}$$ (otherwise $$g(x) \in \mathcal{S}$$ implies $$x = g(g(x)) = g(x)$$, a contradiction), leading to $$f(g(x)) > \frac{1}{g(x)}$$.

Applying these inequalities to $$P(x)$$ gives $$xf(g(x)) + f(x)g(x) < 2$$, which is clearly a contradiction as $$\frac{x}{g(x)} + \frac{g(x)}{x} \geqslant 2$$, e.g., from the AM-GM inequality. Therefore, we must have $$x \in \mathcal{S}$$ for every $$x \in \mathbb{R}^+$$, i.e., $$g(x) = x$$.

Substituting this relationship into the original equation, we obtain $$P(x): \quad xf(x) + f(x)x \leq 2 \implies xf(x) \leq 1 \implies f(x) \leq \frac{1}{x}$$ for every $$x \in \mathbb{R}^+$$. Applying $$yf(y) \leq 1$$ to the equation $$xf(y) + yf(x) > 2$$ (since $$g(x) = x$$) yields $$f(x) > \frac{2}{y} - \frac{x}{y^2}$$, and taking the limit $$y \to x$$ from either side results in $$f(x) \geq \frac{1}{x}$$.

Combining the results, we have $$f(x) \leq \frac{1}{x}$$ and $$f(x) \geq \frac{1}{x}$$, implying $$f(x) = \frac{1}{x}$$ as desired. $$\blacksquare$$

Note: This solution is written more extensively and with more details than necessary for a competition, especially since I include comments at certain points to encourage understanding of the ideas and explain the solution. In practice, this idea would take up only a few lines.