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.

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2022 IMO Problems/Problem 2

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