Let's find all functions $f:(0, \infty) \rightarrow(0, \infty)$ such that for any positive real numbers $x, y$,

$$
f(x) f(y f(x))=f(x+y)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f059ac0d4ad8028bd2d1g-1.jpg?height=54&width=349&top_left_y=282&top_left_x=1598)[^0]


[^0]:    ${ }^{0}$ International Mathematical Competition for University Students
First, we prove that for any $x$, $f(x) \leq 1$. Suppose there exists some $x$ such that $f(x) > 1$. For this $x$, we can find a positive $y$ such that $y f(x) = x + y$; this number is $y = \frac{x}{f(x) - 1}$. Substituting these values into (1),

$$
f(x) f(x+y) = f(x) f(y f(x)) = f(x+y)
$$

from which it follows that $f(x) = 1$, which is a contradiction.

In equation (1), the factor $f(y f(x))$ is always at most 1; this implies that for any $x, y$, $f(x+y) \leq f(x)$, meaning that the function $f$ is monotonically decreasing.

Now we show that the function is either constantly 1 or all its values are less than 1. If for some $x > 0$, $f(x) = 1$, then from the equation $f(2x) = f(x + x) = f(x) f(x f(x)) = 1$. Repeating this step, we get that $f(4x)$, $f(8x)$, etc., are all 1. By monotonicity, the function value must also be 1 at points smaller than these. Therefore, if the function takes the value 1 at least once, it is constantly 1.

The constant 1 function satisfies the functional equation (1). In the following, we consider the case where $f$ takes values less than 1 everywhere. For such functions, we obtain by the previous method that they are strictly monotonically decreasing and thus bijective.

Let $f(1) = a$; for any $t > 0$, substitute into (1) the numbers $x = 1, y = \frac{t}{a}$, and $x = t, y = 1 + \frac{1-a}{a} t$; their sums are equal, so using (1):

$$
\begin{gathered}
f(1) f(t) = f(1) f\left(\frac{t}{a} f(1)\right) = f\left(1 + \frac{t}{a}\right) = \\
= f\left(t + \left(1 + \frac{1-a}{a} t\right)\right) = f(t) f\left(\left(1 + \frac{1-a}{a} t\right) f(t)\right)
\end{gathered}
$$

Dividing by $f(t)$, we get

$$
f(1) = f\left(\left(1 + \frac{1-a}{a} t\right) f(t)\right)
$$

and since $f$ is bijective,

$$
1 = \left(1 + \frac{1-a}{a} t\right) f(t)
$$

from which

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

Introducing the notation $b = \frac{1-a}{a}$, we have $f(t) = \frac{1}{1 + b t}$, where $b > 0$. Substitution can verify that functions of this form are also solutions. The degenerate case $b = 0$ gives the constant 1 function. Therefore, all solutions are:

$$
f(t) = \frac{1}{1 + b t}
$$

where $b \geq 0$.