TITLE: A functional equation with inequality
QUESTION [7 upvotes]: Find all (at least one) functions $f\colon \mathbb{R}\to \mathbb{R}$ (or show there is none), such that 
$$
f(x^3+x)≤x≤f(x^3)+f(x), \quad \text{for all $x\in \mathbb{R}$}.
$$
This is a problem asked by a friend of mine, and I do not know how to proceed. 
Known easy facts:

$f(0)=0$.
It seems the problem can be split to $x>0$ and $x<0$ separately. Let $f_{-}(x)=-f(-x)$ for $x>0$, then $f_{-}(x^3+x)\ge x\ge f_{-}(x^3)+f_{-}(x)$.

REPLY [10 votes]: Question: What is the value of $f(-2)$?
If we let $x=-1$, we see that $f(-2)\le -1$.
Next, let $a$ satisfy $a^3+a=-8$, and note that $a\approx -1.8$.  Then let $x=-2$.  We have:
$$-2\le f(-8)+f(-2)=f(a^3+a)+f(-2)\le a+f(-2)$$
These inequalities give:
$$-2-a\le-1\\-a\le1\\a\ge-1$$
a contradiction.  No such function exists.