How does one solve$\displaystyle 4x \le x^3$?

This is actually a two part question. First, you can't really divide both sides by $\displaystyle x$, since it could be negative (or even zero).

I figure we can do this problem by separating it into three cases. First, let $\displaystyle x=0$ and see if it is a solution (it is). Then let $\displaystyle x>0$ and $\displaystyle x<0$.

When $\displaystyle x>0$, then $\displaystyle 4 \le x^2$. Then we can take the square root of both sides and get $\displaystyle \pm 2 \le x$. If we interpret that as $\displaystyle 2 \le \left| x \right|$ and then write it as $\displaystyle x \le -2$ or $\displaystyle x \ge 2$, we have the correct full answer. But why is that the correct interpretation? How does someone explain that to his daughter?

Last case is to let $\displaystyle x<0$. Then $\displaystyle 4 \ge x^2$ and $\displaystyle \pm 2 \ge x$. Using the same logic as last paragraph, that would indicate that $\displaystyle 2 \ge \left| x \right|$, or that $\displaystyle -2 \le x \le 2$, which is wrong.

What's the correct way to solve$\displaystyle 4x \le x^3$?