Determine if $x=y^3+y+|y|$ is a function We have this relation:
$$x=y^3+y+|y|$$
Problem: Determine if $y$ is a function of $x$.
If there were no $|y|$, I could prove that $y$ is a function of $x$:
$$x_1=x_2 \Rightarrow y_1^3 + y_1 = y_2^3 + y_2 \Rightarrow y_1^3 - y_2^3 + y_1 - y_2 =0 $$
$$\Rightarrow (y_1 - y_2)(y_1^2 + y_1y_2 + y_2^2+1) = 0 \qquad (\mathcal{A})$$
And $y_1^2 + y_1y_2 + y_2^2 + 1$ is always greater than (or equal to) $1$ because:
$$y_1^2 + y_1y_2 + y_2^2 + 1 \ge 1\Longleftrightarrow y_1^2 + y_1y_2 + y_2^2 \ge 0 \Longleftrightarrow 2y_1^2 + 2y_1y_2 + 2y_2^2 \ge 0 \Longleftrightarrow (y_1^2+2y_1y_2+y_2^2)+y_1^2+y_2^2 \ge 0 \Longleftrightarrow (y_1-y_2)^2+y_1^2+y_2^2 \ge 0$$
Based on $(\mathcal{A})$ we can deduct that:
$$
y_1 - y_2 =0 \Rightarrow y_1 = y_2
$$
But this would be true if there were no $|y|$.
Any ideas to solve the problem would be appreciated.

If $y \ge 0$ then $x = y^3 + 2y$  which is strictly increasing on $[0;\infty[$
Else $x = y^3 $ which is strictly increasing on $]-\infty;0[$
$f(y) = y^3+y+|y|$ is continuous and strictly increasing $]-\infty;0[$ and on $[0;\infty[$ therefore strictly increasing on $\mathbb{R}$
So $f$ is injective (and even bijective) and therefore $y$ is a function of $x$
