Understand the problem

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
\[f(x + y) + y \le f(f(f(x)))\]
holds for all $x, y \in \mathbb{R}$.
Source of the problem
Benelux MO 2013
Topic
Functional Equations
Difficulty Level
Easy
Suggested Book
Functional Equations by BJ Venkatachala

Start with hints

Do you really need a hint? Try it first!

Note that the RHS does not contain y. Thus it should be possible to play with different values of y to get rid of the inconvenient f\circ f\circ f.
Taking y=f(f(f(x)))-f(x), we get f(f(f(x)))\le f(x). Now show that f(x)\le f(f(f(x))). Thus $f(f(f(x)))=f(x)$.
Taking x=0, we get f(y)+y\le f(0). Now try to show that f(y)+y\ge f(0).
Taking y=-x we get the desired inequality from hint 3. Thus f(y)+y=f(0)\implies f(y)=f(0)-y for all y.

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