Find all functions from $\mathbb{R}_{>0}$ to $\mathbb{R}_{>0}$ such that:

$$
f(x) f(y)=2 f(x+y f(x))
$$

## Solutions
In general, when dealing with inequalities on the positives (or more generally a restricted domain of definition), it is always interesting to bring out inequalities. Here we start by noting that if $f(x)x$ and $f(z)=2$, we can then choose $y$ such that $x+y f(x)=z$, and we have $f(x) f(y)=2 f(z)=4$. Since $f(x) \geqslant 2$ and $f(y) \geqslant 2$, we must have equality, and thus $f(x)=2$. Finally, we obtain that $f$ is constantly equal to 2, which works.

## 5 The Revenge of Inequalities (Martin & Alexander)

In 2020, after 8 years of absence (which might as well be an eternity), an enemy we no longer feared reappeared in the IMO problems: an inequality problem, a phantom threat.

In the 2021 problem set, confirming our worst fears, a new, particularly daunting inequality problem once again served as problem 2. This marked the definitive advent of the attack of inequalities.

Our goal here is, and those who have seen the prequel to Star Wars will understand, to avoid having to endure the revenge of inequalities, and to prevent our young French Olympians, who are not sufficiently prepared for this exercise, from being forced to bow before the problems.

## Exercises