Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that, for every $x \in \mathbb{R}_{>0}$, there exists a unique $y \in \mathbb{R}_{>0}$ satisfying  
$$ x f(y)+y f(x) \leqslant 2 $$  
(Netherlands)
First we prove that the function \( f(x) = \frac{1}{x} \) satisfies the condition of the problem statement. The AM-GM inequality gives
\[ \frac{x}{y} + \frac{y}{x} \geqslant 2 \]
for every \( x, y > 0 \), with equality if and only if \( x = y \). This means that, for every \( x > 0 \), there exists a unique \( y > 0 \) such that
\[ \frac{x}{y} + \frac{y}{x} \leqslant 2, \]
namely \( y = x \). Let now \( f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0} \) be a function that satisfies the condition of the problem statement. We say that a pair of positive real numbers \( (x, y) \) is \(\operatorname{good}\) if \( x f(y) + y f(x) \leqslant 2 \). Observe that if \( (x, y) \) is good, then so is \( (y, x) \). Lemma 1.0. If \( (x, y) \) is good, then \( x = y \). Proof. Assume that there exist positive real numbers \( x \neq y \) such that \( (x, y) \) is good. The uniqueness assumption says that \( y \) is the unique positive real number such that \( (x, y) \) is good. In particular, \( (x, x) \) is not a good pair. This means that
\[ x f(x) + x f(x) > 2 \]
and thus \( x f(x) > 1 \). Similarly, \( (y, x) \) is a good pair, so \( (y, y) \) is not a good pair, which implies \( y f(y) > 1 \). We apply the AM-GM inequality to obtain
\[ x f(y) + y f(x) \geqslant 2 \sqrt{x f(y) \cdot y f(x)} = 2 \sqrt{x f(x) \cdot y f(y)} > 2. \]
This is a contradiction, since \( (x, y) \) is a good pair. By assumption, for any \( x > 0 \), there always exists a good pair containing \( x \), however Lemma 1 implies that the only good pair that can contain \( x \) is \( (x, x) \), so
\[ x f(x) \leqslant 1 \quad \Longleftrightarrow \quad f(x) \leqslant \frac{1}{x} \]
for every \( x > 0 \). In particular, with \( x = \frac{1}{f(t)} \) for \( t > 0 \), we obtain
\[ \frac{1}{f(t)} \cdot f\left(\frac{1}{f(t)}\right) \leqslant 1. \]
Hence
\[ t \cdot f\left(\frac{1}{f(t)}\right) \leqslant t f(t) \leqslant 1. \]
We claim that \( (t, \frac{1}{f(t)}) \) is a good pair for every \( t > 0 \). Indeed,
\[ t \cdot f\left(\frac{1}{f(t)}\right) + \frac{1}{f(t)} f(t) = t \cdot f\left(\frac{1}{f(t)}\right) + 1 \leqslant 2. \]
Lemma 1 implies that \( t = \frac{1}{f(t)} \Longleftrightarrow f(t) = \frac{1}{t} \) for every \( t > 0 \).