Mongolian Mathematical Olympiad

Answer: $f=1$ and $f=1/x$. It is easy to check that these are solutions, so we prove there are no other solutions.

f = 1 if f is not injective: Suppose $a>b>0$ and $f(a)=f(b)$. Then we have $$f(1+ax)=f(a)f(x+f(a))=f(b)f(x+f(b))=f(1+bx)$$ for any $x\in {\mathbb{R}}_{>0}$. Let $c=a/b>1$ and $x=y/b$, then we get $f(1+cy)=f(1+y)$. Taking $a=1+cy$, $b=1+y$, we get $$f(1+(1+cy)z)=f(1+(1+y)z)$$ for all $z\in {\mathbb{R}}_{>0}$. Now, for $1<t<c$, let $y=\frac{t-1}{c-t}>0$ and $z=\frac{c-t}{c-1}>0$. Then $(1+cy)z=t$ and $(1+y)z=1$, thus $f(1+t)=f(2)$. Moreover, we have

$f(1+c^{n}x)=f(1+x)$ for any $n\ge 1$ by induction, thus increasing $n$, we see that $f(x)=f(2)$ is constant for all $x>2$. Finally, for any $x\in {\mathbb{R}}_{>0}$, let $y=2+1/x$, then $y+f(x)>2$ and $1+xy>2$, thus $f(x)=f(1+xy)/f(y+f(x))=1$.

$f=1/x$ if $f$ is injective: Let $x>1$. For $y=\frac{x-1}{x}>0$, we have $1+xy=x$, thus $f(1-1/x+f(x))=1,$ since $f(x)>0$. Let $d=f(2)-1/2$. Then $f$ injective implies that $f(x)=d+1/x$. For $y=x$, we have $x>1$, $y+f(x)>1$, $1+xy>1$, hence $$\left(d+\frac{1}{x}\right)\left(d+\frac{1}{x+d+\frac{1}{x}}\right)=d+\frac{1}{1+x^2}.$$ Simplifying to a polynomial identity and considering the constant term, we see that $d=0$. It follows that $f(x)=1/x$ for $x>1$. Now let $y=f(x)$. Then $f(2f(x))=f(1+xf(x))/f(x)$ and thus $$f\left(\frac{2}{x}\right)=\frac{x}{1+x\frac{1}{x}}=\frac{x}{2}.$$ Therefore $f(x)=1/x$ for $0<x<2$.