IMO 2022 · Problem 2

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 \[ xf(y)+yf(x)\leq 2. \]

Official text of the problem. Official problem paper (PDF)

Answer

The only such function is \[ \boxed{f(x)=\frac1x\qquad\text{for all }x>0.} \]

Solution

Let \(f\) satisfy the condition. We first prove that, for each \(x>0\), its unique corresponding \(y\) is \(x\) itself.

Fix \(x>0\), and let \(y>0\) satisfy \[ xf(y)+yf(x)\leq 2. \] Suppose \(y\ne x\). By uniqueness for \(x\), the choice \(y=x\) must fail, so \[ xf(x)>1. \] The inequality is symmetric in \(x\) and \(y\). Thus \(x\) is the unique number satisfying the condition for input \(y\). Since \(x\ne y\), the choice of \(y\) itself must fail, giving \[ yf(y)>1. \] But the arithmetic–geometric mean inequality now yields \[ xf(y)+yf(x) \geq 2\sqrt{xf(y)\,yf(x)} =2\sqrt{xf(x)\,yf(y)} >2, \] a contradiction. Consequently, the unique corresponding number is always \(y=x\), and hence \[ xf(x)\leq 1\qquad\text{for every }x>0. \tag{1} \]

We next show that equality always holds in (1). Suppose instead that \(xf(x)<1\) for some \(x>0\), and set \[ y=\frac1{f(x)}. \] Then \(y>x\). Applying (1) at \(y\), we obtain \(f(y)\leq 1/y\), so \[ xf(y)+yf(x) \leq \frac{x}{y}+1 =xf(x)+1 <2. \] Thus this \(y\ne x\) also satisfies the required inequality, contradicting uniqueness. Therefore \(xf(x)=1\) for every \(x>0\), proving the claimed form.

Finally, for \(f(x)=1/x\), \[ xf(y)+yf(x) =\frac{x}{y}+\frac{y}{x} =2+\frac{(x-y)^2}{xy}. \] Since \(xy>0\), this is at most \(2\) exactly when \(y=x\). The required existence and uniqueness therefore hold.

The solution was checked against the official IMO shortlist solutions.