Find the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$ :

$$
f(x+y)+y \leqslant f(f(f(x)))
$$
We start by setting $y=f(f(x))-x$, which gives $f(f(x)) \leqslant x$, and thus the equation yields $f(x+y)+y \leqslant f(x)$. By setting $z=x+y$, we get $f(z)+z \leqslant f(x)+x$. Since this holds for any $x$ and $z$, we must have $f(x)+x$ constant, hence $f(x)=C-x$, which works.