
6. Find all $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$satisfying

$$
f(x) f(y)=f(y) f(x f(y))+\frac{1}{x y}
$$

for any $x, y \in \mathbb{R}^{+}$, where $\mathbb{R}^{+}$denotes the set of positive real numbers.

(Pavel Calábek)


Solution. The formula implies $f(y) \neq 0$ for any $y>0$, thus

$$
f(x f(y))=f(x)-\frac{1}{x y f(y)}
$$

Let us denote $f(1)=a>0$. If we substitute $x=1$, resp. $y=1$ into (1) we get

$$
\begin{gathered}
f(f(y))=f(1)-\frac{1}{y f(y)}=a-\frac{1}{y f(y)} \quad\left(y \in \mathbb{R}^{+}\right) \\
f(a x)=f(x)-\frac{1}{a x} \quad\left(x \in \mathbb{R}^{+}\right)
\end{gathered}
$$

Substituting $x=1$ into (3) yields

$$
f(a)=f(1)-\frac{1}{a}=a-\frac{1}{a}
$$

Choosing $x=a$ in (1) together with (4) gives

$$
f(a f(y))=f(a)-\frac{1}{a y f(y)}=a-\frac{1}{a}-\frac{1}{a y f(y)} \quad\left(y \in \mathbb{R}^{+}\right) .
$$

Now using (3) and (2) we can rewrite the left hand side of the previous equation as

$$
f(a f(y))=f(f(y))-\frac{1}{a f(y)}=a-\frac{1}{y f(y)}-\frac{1}{a f(y)} .
$$

Comparing the right hand sides of the previous two equations we get

$$
f(y)=1+\frac{a-1}{y} \quad\left(y \in \mathbb{R}^{+}\right) .
$$

That is the only possible solutions are of the form (4). Substituting this form into the original equation we find $(a-1)^{2}=1$, which together with the condition $a>0$ gives $a=2$.

Conclusion. The unique solution of the problem is

$$
f(x)=1+\frac{1}{x} .
$$


[^0]:    1 The faces of a dice are numbered similarly: the opposite faces sum up to 7 .

