A functional equation with inequality 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 $f_{-}(x^3+x)\ge x\ge f_{-}(x^3)+f_{-}(x)$.

 A: 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.
A: There is no such function, I believe. To see this, first let $h:\mathbb R\to\mathbb R$ be defined by $h(x)=x^3+x$. This is a sum of two strictly increasing functions and thus strictly increasing. Since it is also unbounded in both directions and continuous, it is bijective, so its inverse $h^{-1}:\mathbb R\to\mathbb R$ exists. (And is also strictly increasing.)
By the required inequalities, this means that for every $x\in\mathbb R$ we have $$f(x)=f(h(h^{-1}(x)))\leq h^{-1}(x).$$ This and the inequalities further imply that $$x\leq f(x^3)+f(x)\leq h^{-1}(x^3)+h^{-1}(x)\tag{*}$$ must hold for all $x\in\mathbb R$. It turns out that this is impossible.
To see this, we must only find an $x\in\mathbb R$ such that $x>h^{-1}(x^3)+h^{-1}(x)$, so that $(*)$ fails. Note that, since $h(-1)=(-1)^3+(-1)=-2$, it follows that $h^{-1}(-2)=-1$. Therefore, $$h^{-1}((-2)^3)+h^{-1}(-2)<h^{-1}(-2)+h^{-1}(-2)=-2,$$ so $(*)$ indeed fails. Therefore no such $f$ exists.
