2022 IMO Problems/Problem 2

Problem

Let denote the set of positive real numbers. Find all functions such that for each , there is exactly one satisfying

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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.

See Also

2022 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
All IMO Problems and Solutions