Bulgarian National Mathematical Olympiad

If $g(x)=1/f(x)$ then $g:{\mathbb{Q}}^{+}\to {\mathbb{R}}^{+}$ and the given equation becomes $$(\ast )g(x+y)g(x)g(y)=g(xy)(g(x)+g(y)).$$ Let $f(1)=c>0$ i.e. $g(1)=\frac{1}{c}$. From () we get $g(x+1)=cg(x)+1$ and hence $g(2)=2$, $g(3)=2c+1$, $g(4)=2c^2+c+1$, $g(5)=2c^3+c^2+c+1$, $g(6)=2c^4+c^3+c^2+c+1$. On the other hand setting $x=2$, $y=3$ in () leads us to $$g(5)g(2)g(3)=g(6)(g(2)+g(3)),$$ which implies $$4c^5-3c^3-c^2-c+1=0\iff (c-1)(c+1)(2c-1)(2c^2+2c+1)=0.$$ If $c=1$ then $g(x+1)=g(x)+1$. By induction $g(n)=n$ for any $n\in \mathbb{N}$ and moreover $g(x+n)=g(x)+n$ for any $x\in {\mathbb{Q}}^{+}$ and $n\in \mathbb{N}$. By setting $y=n$ in (*) we obtain $$(g(x)+n)g(x)n=g(nx)(g(x)+n),$$ i.e. $g(nx)=ng(x)$. Setting $x=p/q$ and $n=q$, where $p,q\in \mathbb{N}$ we obtain $g(x)=x$, for any $x\in {\mathbb{Q}}^{+}$, i.e. $f(x)=1/x$, for any $x\in {\mathbb{Q}}^{+}$.

If $c=\frac{1}{2}$ then $g(x+1)=\frac{1}{2}g(x)+1$. Hence $$g(n)=2\text{and}g(x+n)-2=\frac{g(x)-2}{2^n},x\in {\mathbb{Q}}^{+},n\in \mathbb{N};$$ Setting $y=n$ in $(\ast )$ we obtain $$2g(x+n)g(x)=g(nx)(g(x)+2),$$ and it's sufficient to see that these equations imply $g(x)=2$, for any $x\in {\mathbb{Q}}^{+}$, i.e. $f\equiv 1/2$. Finally, the only functions satisfying the given equality are $f\equiv \frac{1}{2}$ and $f\equiv \frac{1}{x}$.