USAMO Question 2 (via artofproblemsolving.com):
Find all functions $f:(0,\infty)\to(0,\infty)$ such that
$$f\left(x+\frac1y\right)+f\left(y+\frac1z\right)+f\left(z+\frac1x\right)=1$$
for all $x, y, z>0$ with $xyz=1$.
The link redirects to a forum on AoPS. Check USAMO 2's solution. The author claims that $g$ can (uniquely) be extended to an additive function $h$ on all of $\mathbb{R}$. I don't understand why this is true by the explanations below this statement. Please help me understand. Maybe define the function $h$ explicitely. I was thinking of something like
$$h(x)=g\left(x-\left\lfloor x+\frac13 \right\rfloor\right)+3\left\lfloor x+\frac13 \right\rfloor g\left(\frac13 \right), \forall x\in \mathbb R$$ but I don't think this works.
If you can't explain the solution in the link above, but you have a solution to this problem which does not involve analysis (only algebra, and other than Evan Chen's solution / AoPS solutions which are very long and hard to find in a contest), please post it here. It will help. Thank you in advance!
Best Answer
We have $g: (-\frac{1}{3},\frac{1}{3}) \to \mathbb{R}$ with $g(x+y) = g(x)+g(y)$. Define $G: \mathbb{R} \to \mathbb{R}$ by $G(x) = Ng(\frac{x}{N})$ where $N \in \mathbb{N}$ is large enough to ensure $|\frac{x}{N}| < \frac{1}{3}$. To see that the definition does not depend on $N$, i.e. to show $Ng(\frac{x}{N}) = Mg(\frac{x}{M})$ for any $M$ with $|\frac{x}{M}| < \frac{1}{3}$, it suffices to show both are equal to $NMg(\frac{x}{NM})$, which is clear from additivity. Let's show $G(x+y) = G(x)+G(y)$ for $x,y \in \mathbb{R}$. Fix $x,y \in \mathbb{R}$, and take $N$ large so that $|\frac{x}{N}|,|\frac{y}{N}|,|\frac{x+y}{N}| < \frac{1}{3}$; then $G(x+y) = Ng(\frac{x+y}{N})$ and $G(x)+G(y) = Ng(\frac{x}{N})+Ng(\frac{y}{N})$, so just use additivity of $g$. Finally, it is clear that $G$ extends $g$.