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.

Assume that $y>0$. Then it makes $x$ to be $>0$. Now let $x_1=x_2$ and we are on the way to show that the value $y_1$ respect to $x_1$ is the same as $y_2$ for $x_2$. If $y_1\neq y_2$ so $y_2-y_1\neq 0$ and so $$(y_1-y_2)(y_1^2+y_1y_2+y_2^2)=-2(y_1-y_2)$$ reaches us to have a contradiction! For $y<0$ is easy to see what we needs.

