An IMO 2022 Problem

I accidentally encountered this problem when I was surfing facebook. Yeah, I know it sounds crazy that I can study anytime, even during my free time. 🤣 LOL!

Anyway, a few days after I solved it, I found out where it came from. An IMO problem! However, it is not too difficult, but on the contrary, it is quite simple, straightforward, and very interesting when you only need basic knowledge to solve.

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: $xf(y) + yf(x) \leq 2$.

More simply, it can be stated as follows: Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $\forall x > 0, \exists! y > 0, xf(y) + yf(x) \leq 2$.

Next, before discussing how I formed the ideas to solve this problem in detail, why don’t you sit down, take a cup of tea or coffee, and try it yourself? I’m sure it is very fun.

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First of all, this is all knowledge you need to know. Many students in Vietnam are quite familiar with it, so I think it is not hard enough for you.

• Inequality of arithmetic and geometric means: Ref.
• Solving a quadratic inequation.

Now let’s begin!

I don’t remember exactly all ideas I came up with at first. But there was some progress after I tried some trivial functions and found that $f(x) = \frac{1}{x}$ was one of potential solutions for this problem. Thus, if $f(x) = \frac{1}{x}$, then $xf(y) + yf(x)$ will become $\frac{x}{y} + \frac{y}{x}$. This expression is closely related to the number $2$ due to the inequality $\frac{x}{y} + \frac{y}{x} \geq 2, \forall x, y > 0$.

Hmm, what can we do with this new finding? We know that there always exists pairs $(x, y)$ satisfying $xf(y) + yf(x) \leq 2$ with $x, y > 0$. From $xf(y) + yf(x) \leq 2$, we have:

\[x\left(f(y) - \frac{1}{y}\right) + y\left(f(x) - \frac{1}{x}\right) \leq - \frac{x}{y} - \frac{y}{x} + 2 \leq 0\]

That means $f(x) > \frac{1}{x}, \forall x > 0$ cannot be true because there are no pairs $(x, y)$ satisfying requirements. Therefore, some $x$ may satisfy $f(x) > \frac{1}{x}$ while other $x$ may satisfy $f(x) \leq \frac{1}{x}$.

The next question is whether we can improve the above result, i.e. there doesn’t exist $x$ satisfying $f(x) > \frac{1}{x}$. Why don’t we assume that $x_0$ is a such value, i.e. $f(x_0) > \frac{1}{x_0}$? If true, there is exactly one $y_0$ satisfying the inequation. Then, we infer that $f(y_0) \leq \frac{1}{y_0}$, so $x_0 \neq y_0$. However, if we choose the value $y_0$ for $x$, we have two distinct values of $y$ satisfying the inequation, $x_0$ and $y_0$. This is contrary to the hypothesis, and we conclude that there are no values of $x$ satisfying $f(x) > \frac{1}{x}$, i.e. $f(x) \leq \frac{1}{x}, \forall x > 0$.

What about $f(x) < \frac{1}{x}$? Can we treat it the same as $f(x) > \frac{1}{x}$, i.e. there doesn’t exist $x$ satisfying $f(x) < \frac{1}{x}$, which implies $f(x) = \frac{1}{x}, \forall x > 0$? Luckily, this is possible, but my way is slightly not natural. Firstly, we assume that $y_1$ satisfies $f(y_1) < \frac{1}{y_1}$. Secondly, we have $f(x) \leq \frac{1}{x}, \forall x > 0$, so $xf(y) + yf(x) \leq xf(y) + \frac{y}{x}, \forall x > 0$. Thirdly, we can show that for each $x > 0$, $x$ is a suitable value of $y$ satisfying the inequation due to $f(x) \leq \frac{1}{x}, \forall x > 0$. Hence, from the above three findings, if we can choose an infinite number of positive values for $x$ such that $x \neq y_1$ and $y_1$ is also a suitable value of $y$ satisfying the inequation, the assumption will be fail because it will lead to a contradiction with the hypothesis. This can be viable if we choose $x > 0$ satisfying $x \neq y_1$ and $xf(y_1) + \frac{y_1}{x} \leq 2$. For all $x > 0$, we have:

\[\begin{aligned} & xf(y_1) + \frac{y_1}{x} \leq 2 \\ \iff & x^2 - \frac{2}{f(y_1)}x + \frac{y_1}{f(y_1)} \leq 0 \\ \iff & \left(x - \frac{1}{f(y_1)}\right)^2 - \frac{1 - y_1f(y_1)}{f(y_1)^2} \leq 0 \\ \iff & \left(x - \frac{1 - \sqrt{1 - y_1f(y_1)}}{f(y_1)}\right)\left(x - \frac{1 + \sqrt{1 - y_1f(y_1)}}{f(y_1)}\right) \leq 0 \\ \end{aligned}\]

Then, all values in $\left[\frac{1 - \sqrt{1 - y_1f(y_1)}}{f(y_1)}, \frac{1 + \sqrt{1 - y_1f(y_1)}}{f(y_1)}\right] \setminus \{ y_1 \}$ satisfy the above requirements of $x$. Therefore, the assumption is false, so $f(x) = \frac{1}{x}, \forall x > 0$.

So that’s the way I solved this problem. Thank you for reading to here and have a nice day :)