# Finding $f$ such that $f(x+\frac1y)+f(y+\frac1z)+f(z+\frac1x)=1$ for positive $x$, $y$, $z$ with $xyz=1$. Explain existing answer.

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!

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 $$g$$.
We can fix the extension part of the solution as follows. Put $$U=\left( -\tfrac 13, \tfrac 23\right)$$.
We claim that for each $$x_1,\dots, x_k\in U$$ with $$x_1+\dots+x_k=0$$ we have $$g(x_1)+\dots+g(x_k)=0$$. Let’s prove this claim by induction with respect to $$k$$. For $$k\le 3$$ the claim is given. Assume that the claim is proved for each $$k\le n\ge 3$$. Let $$x_1,\dots, x_{n+1}\in U$$ with $$x_1+\dots+x_{n+1}=0$$. Without loss of generality we can assume that $$x_1\le 0\le x_2$$, so $$x_1+x_2\in [x_1, x_2]\subset U$$. By the inductive hypothesis we have $$g(x_1+x_2)+g(x_3)+\dots+g(x_n)=0,$$ so it remains to prove that $$g(x_1+x_2)=g(x_1)+g(x_2)$$. It is easy to see that $$-\tfrac{x_1+x_2}2\in U$$, so $$g(x_1+x_2)+2g\left(-\tfrac{x_1+x_2}2\right)=0.$$ Similarly we have $$g(x_1)+2g\left(-\tfrac{x_1}2\right)=0\mbox{ and }g(x_2)+2g\left(-\tfrac{x_2}2\right)=0.$$ Moreover, $$g\left(-\tfrac{x_1+x_2}2\right)+ g\left(\tfrac{x_1}2\right)+ g\left(\tfrac{x_2}2\right)=0,$$
$$g\left(\tfrac{x_1}2\right)+ g\left(-\tfrac{x_1}2\right)=0,\mbox{ and } g\left(\tfrac{x_2}2\right)+ g\left(-\tfrac{x_1}2\right)=0.$$ It follows $$g(x_1+x_2)=$$ $$-2g\left(-\frac{x_1+x_2}2\right)=2 g\left(\frac{x_1}2\right)+2g\left(\frac{x_2}2\right)=-2 g\left(-\frac{x_1}2\right)-2g\left(-\frac{2}2\right)=$$ $$g(x_1)+g(x_2).$$
Let $$x\in\Bbb R$$ be any number, $$x=x_1+\dots+x_n$$ and $$x=x’_1+\dots+x’_m$$ be two representations of $$x$$ with $$x_1,\dots, x_n, x’_1,\dots, x’_m\in U$$. Then $$\pm \tfrac {x_i}2$$ and $$\pm \tfrac {x’_j}2$$ belong to $$U$$ for each $$i$$ and $$j$$. By the claim we have $$g(x_1)+\dots+g(x_n)=$$ $$-2\left(g\left(-\frac{x_1}2\right)+\dots+ g\left(-\frac{x_n}2\right) \right)=$$ $$2\left(g\left(\frac{x’_1}2\right)+\dots+ g\left(\frac{x’_m}2\right) \right)=$$ $$-2\left(g\left(-\frac{x’_1}2\right)+\dots+ g\left(-\frac{x’_m}2\right) \right)=$$ $$g(x’_1)+\dots+g(x’_n).$$
Put $$h(x)=g(x_1)+\dots+ g(x_n)$$. The definition of $$h(x)$$ implies that $$h$$ is additive and an extension of $$g$$. The uniqueness of such an $$h$$ follows from its additivity and the claim, but I guess it is not needed for the solution, since existence of any additive extension of $$g$$ on $$\Bbb R$$ implies $$g(x)=kx$$ for some $$k\in\left[-\tfrac 12,1\right]$$.