Find the functions $f: \mathbb{R} \leftarrow \mathbb{R}$ such that:

$$
f(x+y)+y \leq f(f(f(x)))
$$
Let $y=f(f(x))-x$. Then we have:

$$
f(f(f(x)))+f(f(x))-x \leq 0 \Longrightarrow f(f(x)) \leq x
$$

Let $z=x+y$. It follows that:

$$
f(z)+z-x \leq f(f(f(x)))
$$

From the above:

$$
f(z)+z-x \leq f(x) \Longleftrightarrow f(z)+z \leq f(x)+x
$$

This is true for all $x, z$, so $f(x)+x$ is constant. Thus, $f(x)=-x+C$. Conversely, it is easy to verify that these functions are solutions.