Find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that
$$
f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right)+f\left(z+\frac{1}{x}\right)=1
$$
for all $x, y, z>0$ with $x y z=1$.
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 $f(t)=g\left(\frac{1}{t+1}\right)$, equivalently $g(s)=f(1 / s-1)$. Thus $g:(0,1) \rightarrow(0,1)$ which satisfies $\sum_{\text {cyc }} g\left(\frac{a}{a+b+c}\right)=1$, or equivalently
$$
g(a)+g(b)+g(c)=1 \quad \forall a+b+c=1 .
$$

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).
Claim - The function $g$ is linear.

Proof. This takes several steps, all of which are technical. We begin by proving $g$ is linear over $[1 / 8,3 / 8]$.
- First, whenever $a+b \leq 1$ 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 $(0,1 / 2)$.
- Define $h:[0,1] \rightarrow \mathbb{R}$ by $h(t)=g\left(\frac{2 t+1}{8}\right)-(1-t) \cdot g(1 / 8)-t \cdot g(3 / 8)$, 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.
- Since
$$
h(t)=h(t)+h(1 / 2)=2 h(t / 2+1 / 4)=h(t+1 / 2)+h(0)=h(t+1 / 2)
$$
for any $t<1 / 2$, we find $h$ is periodic modulo $1 / 2$. It follows one can extend $\widetilde{h}$ by
$$
\widetilde{h}: \mathbb{R} \rightarrow \mathbb{R} \quad \text { by } \quad \widetilde{h}(t)=h(t-\lfloor t\rfloor)
$$
and still satisfy Jensen's functional equation. Because $\widetilde{h}(0)=0$, it's well-known this implies $\widetilde{h}$ is additive (because $\widetilde{h}(x+y)=2 \widetilde{h}((x+y) / 2)=\widetilde{h}(x)+\widetilde{h}(y)$ for any real numbers $x$ to $y$ ).
But $\widetilde{h}$ is bounded below on $[0,1]$ since $g \geq 0$, and since $\widetilde{h}$ is also additive, it follows (well-known) that $\widetilde{h}$ is linear. Thus $h$ is the zero function. So, the function $g$ is linear over $[1 / 8,3 / 8]$; thus we may write $g(x)=k x+\ell$, valid for $1 / 8 \leq x \leq 3 / 8$.

Since $3 g(1 / 3)=1$, it follows $k+3 \ell=1$.
For $0<x<1 / 8$ we have $g(x)=2 g(0.15)-g(0.3-x)=2(0.15 k+\ell)-(k(0.3-x)+\ell)=$ $k x+\ell$, so $g$ is linear over $(0,3 / 8)$ as well. Finally, for $3 / 8<x<1$, we use the given equation
$$
1=g\left(\frac{1-x}{2}\right)+g\left(\frac{1-x}{2}\right)+g(x) \Longrightarrow g(x)=1-2\left(k \cdot \frac{1-x}{2}+\ell\right)=k x+\ell
$$
since $\frac{1-x}{2}<\frac{5}{16}<\frac{3}{8}$. Thus $g$ is linear over all.
Putting this back in, we deduce that $g(x)=k x+\frac{1-k}{3}$ 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.