Difference between revisions of "2022 IMO Problems/Problem 2"
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| + | 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 \( y = x \). | ||
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| + | 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. | ||
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| + | 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 \( y \neq g(x) \). | ||
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| + | 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). | ||
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| + | 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. | ||
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| + | 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 \( f(g(x)) > \frac{1}{g(x)} \). | ||
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| + | 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., \( g(x) = x \). | ||
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| + | 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 \( f(x) \geq \frac{1}{x} \). | ||
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| + | 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. \(\blacksquare\) | ||
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| + | 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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| + | ==See Also== | ||
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| + | {{IMO box|year=2022|num-b=1|num-a=3}} | ||
Latest revision as of 11:41, 20 December 2023
Problem
Let 
denote the set of positive real numbers. Find all functions 
such that for each 
, there is exactly one 
satisfying
![\[xf (y) + yf (x) \le 2\]](http://latex.artofproblemsolving.com/a/5/2/a52c8066dc22101920e4ab14534ff3bb74d149fb.png)
.
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 | ||
