Find all functions $f : (0, ∞) → (0, ∞)$ such that $f(x +\frac{1}{y})+ f(y +\frac{1}{z})+ f(z + \frac{1}{x})= 1$ for all $x, y, z > 0$ with $xyz = 1$.
Alright so my main question is that i first saw evan's solution which was very tricky and random. After that i decided that should infact have a look at a few more solutions to undesrtand whats goin on. Then on yt i stumbled upon Osman nal's solution to the same question (i don't know if it is his own or whether he got it from somewhere). They both have the smae first few steps after which they diverge. Anyway what evan has done looks so complicated and random while at that exact point Osman had made a substitution which pretty much trivialised the question by turning it into a cauchy. I shall provide a very brief outline below.
- Use classic inequality trick of removing the $xyz=1$ constraint by setting $x=\frac{a}{b}$ and so on.
2.Guess a solution at this point ( which is fairly trivial to do and comes out to be $f(x)=\frac{1}{x+1}$.
- Define $g(\frac{1}{t+1})=f(t)$ which then has domain and range both as $(0,1)$. We get a functional equation $g(a)+g(b)+g(c)=1$ with $a+b+c=1$.
Now this is where Evan and Osman make different moves.
https://web.evanchen.cc/exams/USAMO-2018-notes.pdf
This the link to his solution notes for USAMO 2018 and the question is P2 ( sorry for not providing the solution here it is pretty complicated and i would then have to format the entire thing.)
What osman does is as follows:
4.Now define a new function $h(x)=g(x+\frac{1}{3})-\frac{1}{3}$. In other words shifting $g(x)$ by a third.
- Thus our new FE is $h(a)+h(b)+h(c)=0$ with $a+b+c=1$.
6 Simply do a bit of plug and chug to obtain that $h$ is odd, $h(0)=0$, and $h(x)+h(y)=h(x+y)$, which means $h$ is additive.
Since it is bounded on an interval, we deduce that $h(x)=kx$ for some constant $k$, by Cauchy's FE.
Plug this into the original equations and get the final answer.
The thing is evan actually managed to place a constraint on what $k$ should be which is something Osman did do. Now his solution is much neater and appealing but that looks like a flaw to me. What i wish to know is whether his solution is correct or not. This curiousity arises since i really trust Evan's genius and that he would have infact taken the same path had been truly correct and better. Please explain. Thanks.
Osman's solution is slightly incorrect.
(In his video, which I merely skimmed. I'm assuming that these statements are true, which it seems like they are. )
He showed that
From there, he concluded that $ h(x) = c x$, but did not provide any restrictions on $k$. However, it is obvious that not all $c$ work due to the range. From condition 1, we clearly have $ c \in [ - \frac{1}{2} , 1 ]$.
Alternatively, once you arrive at "All possible solutions are of the form $f(x) = \frac{c}{1+x} + \frac{1-c}{3}$", one has to ensure that this satisfies the conditions of the problem - specifically the positive range. From there, we get restrictions on $c$.