I found that
Given two real numbers $s,p$, there exist three real numbers $x,y,z$ such that $x+y+z=s$ and $xyz=p$.
This is easy to prove when $p=0$: pick $x=s$ and $y=z=0$.
To prove the case $p \neq 0$, I consider $z \neq 0$ as a parameter in the quadratic equation $$X^2-(s-z)X+\frac{p}{z}=0$$
The discriminant equals $$\Delta = (s-z)^2 - \frac{4p}{z} \to + \infty$$ as $z \to + \infty$. Thus, for $z$ large enough $\Delta >0$, so that the quadratic equation has two solutions $x,y$ satisfying $$x+y = s-z \ \ ; \ \ xy= \frac{p}{z}$$ as we wanted to prove.
Now, my question is:
if $s,p$ are rational, can we find rational $x,y,z$ such that $x+y+z=s$ and $xyz=p$?
This relies on finding $z \in \Bbb{Q}^*$ such that $\Delta = (s-z)^2 - \frac{4p}{z}$ is the square of a rational number. However, here I got stuck.