From the IMO shortlist:
We denote by $\mathbb{R}^+$ the set of all positive real numbers.
Find all functions $f: \mathbb R^+\rightarrow\mathbb R^+$ which have the property:
$$f(x)f(y)=2f(x+yf(x))$$
for all positive real numbers $x$ and $y$.
$\textbf{My progress: }$At first assume $f$ is not injective.
Then there exist $z,y$ such that $f(z)=f(x)$ assume WLOG $z >x$.Now,We can choose some appropriate $y$ such that $x+yf(x)=z$.
Then it follows that $f(y)=2$ which means $2$ has an inverse. Now, substituting inverse of $2$ in place of $y$ from which we could get a bunch of values for which the function assumes the same value. I then thought of somehow proving this would show the function is constant which I failed to do.
The case with $f$ injective is pretty easy.interchaging $x,y$ we get, $f(x+yf(x))=f(y+xf(y))$ Using injectivity we could easily deduce from here that $f$ is linear and get a contradiction.
But I cannot do the first case. I do believe that $f(x)=2$ is the only function that works.
Any kind of hint or solution is appreciated.
Best Answer
Like you wrote, we can show that $f$ is not injective and there is some $a$ such that $f(a) = 2$. Suppose towards a contradiction that there exists a $b$ with $f( b) < 2$.
$P(a, x)$ gives $f(x) = f(2x + a)$ so we can find a $c > a$, with $f(c ) = f(b)$ Let $y \in \mathbb{R}^+$ such that $$c + f(c)y = 2y + a$$
Then $P(c, y)$ gives $f(c) = 2$ which is a contradiction.
Similarly, suppose that there exists a $b$ with $f(b ) > 2$. Then we can find a $c < a$ with $f(c) = f(b)$ and a $y$ with $$c + f(c)y = 2y + a$$ and $P(c,y)$ gives the same contradiction.