Let $k$ be a real number. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$,

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

(ITYM 2009)
Here is a brief summary of the solution. The complete version can be found on the internet (written solution by the French team 2 for problem 2 of ITYM 2009).

We distinguish two cases.

- If $f(0)=0$, we find that $f \circ f$ is zero, and it quickly follows that $f=0$ if $k=0$. If $k \neq 0$, it is more difficult to show that $f$ is the zero function, but we manage to do so.
- If $f(0) \neq 0$, let $f(0)=a$. With $y=0$ in the equation, we obtain

$$
f(f(x)+a)=a x
$$

so the function is injective. This allows us to inject, for example, $x=f(y)+a$ into the previous relation. It follows that

$$
f(a y+a)=a(f(y)+a)
$$

After some manipulations, we get $k=2$ and $a=\frac{1}{4}$. We then focus on the sets $A=\left\{x \in \mathbb{R}, f(x)=\frac{1}{4}-\frac{x}{2}\right\}$ and $B=\left\{x \in \mathbb{R}, f(x)=2 x^{2}\right\}$, and finally show that $A=\mathbb{R}$.

## Monday, June 25 morning: inequalities, Jean-François Martin

How to handle inequalities when you don't know how to handle inequalities?

One way is to use analysis. Let's illustrate this statement with an extremely simple (and highly caricatural) example.