IMC 2000 · Day 2 · P11

First, if we assume that $f(x) > 1$ for some $x \in \mathbb{R}^+$, setting $y = \dfrac{x}{f(x) - 1}$ gives the contradiction $f(x) = 1$. Hence $f(x) \le 1$ for each $x \in \mathbb{R}^+$, which implies that $f$ is a decreasing function.

If $f(x) = 1$ for some $x \in \mathbb{R}^+$, then $f(x + y) = f(y)$ for each $y \in \mathbb{R}^+$, and by the monotonicity of $f$ it follows that $f \equiv 1$.

Let now $f(x) < 1$ for each $x \in \mathbb{R}^+$. Then $f$ is strictly decreasing function, in particular injective. By the equalities $$\begin{align*} f(x) f(y f(x)) &= f(x + y) = \\ &= f \bigl( y f(x) + x + y (1 - f(x)) \bigr) = f(y f(x)) f \Bigl( \bigl( x + y(1 - f(x)) \bigr) f(y f(x)) \Bigr) \end{align*}$$ we obtain that $x = \bigl( x + y(1 - f(x)) \bigr) f(y f(x))$. Setting $x = 1$, $z = x f(1)$ and $a = \dfrac{1 - f(1)}{f(1)}$, we get $f(z) = \dfrac{1}{1 + az}$.

Combining the two cases, we conclude that $f(x) = \dfrac{1}{1 + ax}$ for each $x \in \mathbb{R}^+$, where $a \ge 0$. Conversely, a direct verification shows that the functions of this form satisfy the initial equality.

As in the first solution we get that $f$ is a decreasing function, in particular differentiable almost everywhere. Write the initial equality in the form $$\frac{f(x+y) - f(x)}{y} = f^2(x)\, \frac{f(y f(x)) - 1}{y f(x)}.$$ It follows that if $f$ is differentiable at the point $x \in \mathbb{R}^+$, then there exists the limit $\lim\limits_{z \to 0+} \dfrac{f(z) - 1}{z} =: -a$. Therefore $f'(x) = -a f^2(x)$ for each $x \in \mathbb{R}^+$, i.e. $\left( \dfrac{1}{f(x)} \right)' = a$, which means that $f(x) = \dfrac{1}{ax + b}$. Substituting in the initial relaton, we find that $b = 1$ and $a \ge 0$.