Problem: Find all functions $f \colon (0,\infty) \to (0,\infty)$ such that
\[
  f\left( x+\frac 1y \right)
  + f\left( y+\frac 1z \right)
  + f\left( z+\frac 1x \right)
  = 1
\]
for all $x,y,z > 0$ with $xyz = 1$.

Solution: The main part of the problem is to show all solutions are linear.
As always, let $x=b/c$, $y=c/a$, $z=a/b$
(classical inequality trick).
Then the problem becomes
\[ \sum_{\text{cyc}} f\left( \frac{b+c}{a} \right) = 1. \]
Let $f(t) = g(\frac{1}{t+1})$,
equivalently $g(s) = f(1/s-1)$.
Thus $g \colon (0,1) \to (0,1)$
which satisfies $\sum_{\text{cyc}} g\left( \frac{a}{a+b+c} \right) = 1$,
or equivalently
\[ \boxed{g(a) + g(b) + g(c) = 1} \qquad \forall a+b+c=1. \]

The rest of the solution is dedicated to solving this
equivalent functional equation in $g$.
It is a lot of technical details and I will only outline them
(with apologies to the contestants who didn't have that luxury).

\begin{claim*}
  The function $g$ is linear.
\end{claim*}
\begin{proof}
  This takes several steps, all of which are technical.
  We begin by proving $g$ is linear over $[1/8, 3/8]$.
  \begin{itemize}
    \ii First, whenever $a+b \le 1$ we have
    \[ 1 - g(1-(a+b)) = g(a) + g(b) = 2 g\left( \frac{a+b}{2} \right). \]
    Hence $g$ obeys Jensen's functional equation over $(0,1/2)$.
    \ii Define $h \colon [0,1] \to \RR$
    by $h(t) = g(\frac{2t+1}{8}) - (1-t) \cdot g(1/8) - t \cdot g(3/8)$,
    then $h$ satisfies Jensen's functional equation too over $[0,1]$.
    We have also arranged that $h(0) = h(1) = 0$, hence $h(1/2) = 0$ as well.
    \ii Since
    \[ h(t) = h(t) + h(1/2) = 2h(t/2+1/4)
      = h(t+1/2) + h(0) = h(t+1/2) \]
    for any $t < 1/2$, we find $h$ is periodic modulo $1/2$.
    It follows one can extend $\wt h$ by
    \[ \wt h \colon \RR \to \RR \qquad\text{by}\qquad
      \wt h(t) = h(t - \left\lfloor t \right\rfloor) \]
    and still satisfy Jensen's functional equation.
    Because $\wt h(0) = 0$, it's well-known this implies $\wt h$ is additive
    (because $\wt h(x+y) = 2\wt h\left( (x+y)/2 \right) = \wt h(x) + \wt h(y)$
    for any real numbers $x$ to $y$).
  \end{itemize}
  But $\wt h$ is bounded below on $[0,1]$ since $g \ge 0$,
  and since $\wt h$ is also additive,
  it follows (well-known) that $\wt h$ is linear.
  Thus $h$ is the zero function.
  So, the function $g$ is linear over $[1/8,3/8]$;
  thus we may write $g(x) = kx + \ell$, valid for $1/8 \le x \le 3/8$.

  Since $3g(1/3) = 1$, it follows $k + 3\ell = 1$.

  For $0 < x < 1/8$ we have
  $g(x) = 2g(0.15) - g(0.3-x) = 2(0.15k+\ell) - (k(0.3-x)+\ell) = kx+\ell$,
  so $g$ is linear over $(0,3/8)$ as well.
  Finally, for $3/8 < x < 1$, we use the given equation
  \[
    1 = g\left( \frac{1-x}{2} \right)
    + g\left( \frac{1-x}{2} \right)
    + g(x)
    \implies g(x) = 1 - 2\left( k \cdot \frac{1-x}{2} + \ell \right)
    = kx+\ell
  \]
  since $\frac{1-x}{2} < \frac{5}{16} < \frac{3}{8}$.
  Thus $g$ is linear over all.
\end{proof}

Putting this back in,
we deduce that $g(x) = kx + \frac{1-k}{3}$ for some $k \in [-1/2,1]$,
and so \[ f(x) = \frac{k}{x+1} + \frac{1-k}{3} \]
for some $k \in [-1/2,1]$.
All such functions work.