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# 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 $x, y \in \mathbb{R}$.
Benelux MO 2013
##### Topic
Functional Equations
Easy
##### Suggested Book
Functional Equations by BJ Venkatachala

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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