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So this goes again in the bad spirit of some of the late international mathematical competitions - problems asked in the contest appeared not long ago in some widely available place. MEMO 2012 Problem 1 = some problem posted on AoPS in 2009. Way to go, lads ...

Mavropnevma you are extremely unjust, Aops is huge and I don't actually see how and why would anybody try to find their problem here. I agree that it is bad practice to put an already seen problem on a competition but there has been tons of posted problems lately, especially when people just post their idea and don't relate it to some competition and it is quite possible that something will come out which has already appeared. I can guarantee to you that author of this problem didn't have any idea about the prior post and if he had this problem wouldn't have been proposed. As well as there was exactly one person who solved this problem in my opinion there weren't too many people who actually saw this problem before http://www.imosuisse.ch/skripte/MEMO/memo2012_individual.pdf As well this is quite short and simple functional equation but a bit different so it is quite easy for to people to get to it independently. As well you should bear in mind that Memo consists of 10 countries so there are 10 people choosing problems so it is quite hard to identify if some problem has already been seen or no. I mean if among 100 people who select Imo problems nobody saw that Imo 2011 problem 3 is practically the same as Isl 2009 a5. and what is even worse same person is author of both of these. And nobody said anything about this. I agree that everybody should try not to propose an alreday seen problem and even more try to stop them from coming on the competition but you should be more considerate before you accuse people it was done on purpose (I am sorry if you didn't mean this but it sounds like that to me).

Well-known trick - set $x+f(y)=xy+1$, which is possible if $\frac{f(y)-1}{y-1} > 0$. This leads us to $y=1$. Therefore, if $y \neq 1$, we have $(f(y)-1)(y-1) \le 0$. In other words, $f(y) \le 1$ if $y \ge 1$ and $f(y) \ge 1$ if $y \le 1$. Now what we do is assume $y>1$ and let $xy+1=y$, which is definitely possible. Plug that and we have $yf(y)=f(1-\frac{1}{y}+f(y))$. From here, $f(y) > \frac{1}{y}$ and $f(y) < \frac{1}{y}$ both lead to contradictions. So we have $f(y)=\frac{1}{y}$ for $y > 1$. Since $xy+1 > 1$, we have $f(x+f(y)) = \frac{y}{xy+1}$. Let $y>1$, to have $f(x+\frac{1}{y}) = \frac{1}{x+ \frac{1}{y}}$. We can represent any $\epsilon \le 1$ as $x+\frac{1}{y}$ with $x>0$ and $y>1$, so $f(x) = \frac{1}{x}$.

$P(x,1)$ $$f(x+f(1))=f(x+1)$$$c=f(1)-1$ $$f(x)=f(x+c)=f(x+nc)$$$P(n,y)-P(n,y+c)$ $$y=y+c$$$$f(1)=1$$$\frac{f(y)-1}{y-1} > 0$ $y=1$ contradiction. for $y \neq 1$ $(f(y)-1)(y-1) \le 0$., $f(y) \le 1$ if $y \ge 1$ and $f(y) \ge 1$ if $y \le 1$. if exist $a>1$ $f(a)a>1$ $P(\frac{a-1}{a},a)$ $$f(\frac{a-1}{a}+f(a))=af(a)>1$$$$\frac{a-1}{a}+f(a)<1$$$$f(a)a<1$$contradiction. similarity $af(a)<1$ contradiction for $x>1$ $f(x)=\frac{1}{x}$ all $x>0$ $f(x)=\frac{1}{x}$ Answer: $f(x)=\frac{1}{x}$

Lovely problem. Let $P(x,y)$ denotes assertion of given functional equation. Note that $P(\frac{1-f(y)}{1-y},y)$ gives us that $y=1$, which is contradiction. This immediately implies that for $y>1$ ,we have $f(y)<1$ and for $y<1$, we have $f(y)>1$. Claim: $f(1)=1$ Proof: Consideing $P(x,1)$ gives us: $$ f(x+f(1)) = f(x+1)$$This means that function $f(x)$ is periodic with period $T= | f(1) -1|$. Consider number $\epsilon <1$, then we have: $$ f(\epsilon) = f(\epsilon+nT) $$By picking $n$ large enough we can guarantee that $\epsilon+nT>1$ ,while $\epsilon <1$. This gives us contradiction with the beforehand established result, unless $f(1) =1$. Thus the claim is proved. Now the key step is as follows. For $y>1$ we consider $P(1 -\frac{1}{y},y)$: $$ f(1- \frac{1}{y}+ f(y)) =yf(y) $$. If $f(y)>\frac{1}{y}$, we have: $$ 1>f(1- \frac{1}{y}+ f(y)) =yf(y) \implies \frac{1}{y} >f(y) $$. which is contradiction. Similarly, we can deal with the case $f(y)<\frac{1}{y}$. This allows us to conclude that $f(y)=\frac{1}{y}$ for every $y \ge 1$. Now looking at $P(1,y)$ tells us: $$ f(1+f(y)) = yf(y+1) \implies \frac{1}{1+f(y)}= \frac{y}{y+1} \implies f(y) =\frac{1}{y}$$It is easy to verify that this function indeed works.

Kimchiks926 wrote: Lovely problem. Let $P(x,y)$ denotes assertion of given functional equation. Note that $P(\frac{1-f(y)}{1-y},y)$ gives us that $y=1$, which is contradiction. This immediately implies that for $y>1$ ,we have $f(y)<1$ and for $y<1$, we have $f(y)>1$. Claim: $f(1)=1$ Proof: Consideing $P(x,1)$ gives us: $$ f(x+f(1)) = f(x+1)$$This means that function $f(x)$ is periodic with period $T= | f(1) -1|$. Consider number $\epsilon <1$, then we have: $$ f(\epsilon) = f(\epsilon+nT) $$By picking $n$ large enough we can guarantee that $\epsilon+nT>1$ ,while $\epsilon <1$. This gives us contradiction with the beforehand established result, unless $f(1) =1$. Thus the claim is proved. Now the key step is as follows. For $y>1$ we consider $P(1 -\frac{1}{y},y)$: $$ f(1- \frac{1}{y}+ f(y)) =yf(y) $$. If $f(y)>\frac{1}{y}$, we have: $$ 1>f(1- \frac{1}{y}+ f(y)) =yf(y) \implies \frac{1}{y} >f(y) $$. which is contradiction. Similarly, we can deal with the case $f(y)<\frac{1}{y}$. This allows us to conclude that $f(y)=\frac{1}{y}$ for every $y \ge 1$. Now looking at $P(1,y)$ tells us: $$ f(1+f(y)) = yf(y+1) \implies \frac{1}{1+f(y)}= \frac{y}{y+1} \implies f(y) =\frac{1}{y}$$It is easy to verify that this function indeed works. Your solution is incomplete since your last assumption is wrong, $\frac{1}{y}<f(y)$ implies $f\left(1-\frac{1}{y}+f(y)\right)=yf(y)>1$, which is just circular.