Problem: Find all functions $f \colon \RR^+ \to \RR^+$ such that for each $x \in \RR^+$,
there is exactly one $y \in \RR^+$ satisfying \[ xf(y)+yf(x) \leq 2. \]

Solution: The answer is $f(x) \equiv 1/x$ which obviously works (here $y=x$).

For the converse, assume we have $f$ such that
each $x \in \RR^+$ has a \emph{friend} $y$ with $xf(y)+yf(x)\le2$.
By symmetry $y$ is also the friend of $x$.

\begin{claim*}
  In fact every number is its own friend.
\end{claim*}
\begin{proof}
  Assume for contradiction $a \neq b$ are friends.
  Then we know that  $af(a) + af(a) > 2 \implies f(a) > \frac 1a$.
  Analogously, $f(b) > \frac 1b$.
  However, we then get
  \[ 2 \ge a f(b) + b f(a) > \frac ab + \frac ba \overset{\text{AMGM}}{\ge} 2 \]
  which is impossible.
\end{proof}

The problem condition now simplifies to saying
\[ f(x) \le \frac1x \text{ for all $x$}, \qquad
  xf(y) + yf(x) > 2 \text{ for all $x \neq y$}. \]
In particular, for any $x>0$ and $\eps > 0$ we have
\begin{align*}
  2 &< xf(x+\eps) + (x+\eps)f(x) \le \frac{x}{x+\eps} + (x+\eps) f(x) \\
  \implies f(x) &> \frac{x+2\eps}{(x+\eps)^2}
  = \frac{1}{x + \frac{\eps^2}{x+2\eps}}.
\end{align*}
Since this holds for all $\eps > 0$ this forces $f(x) \ge \frac1x$ as well.
We're done.

\begin{remark*}
  Alternatively, instead of using $x+\eps$,
  it also works to consider $y = \frac{1}{f(x)}$.
  For such a $y$, we would have
  \[ xf\left( \frac{1}{f(x)} \right) + \frac{1}{f(x)} \cdot f(x)
    = xf\left( \frac{1}{f(x)} \right) + 1
    \leq x \cdot f(x) + 1 \leq 1 + 1 = 2 \]
  which gives a similar contradiction.
\end{remark*}