Is $|x^3-1|\ge1-x$ true for all real numbers? What is the solution set? I have to solve following inequality 
$$|x^3-1|\ge1-x$$
my try: 
Case 1: let $x\ge 1$, 
$$x^3-1\ge1-x\iff (x-1)(x^2+x+2)\ge 0$$
but $x^2+x+2>0$ for all $x\in R$, hence 
$$x-1\ge 0  \iff x\ge 1$$
which is true. 
Case-2: let $x\le 1$, 
$$-(x^3-1)\ge1-x\iff x(x-1)(x+1)\le 0\iff x\in(-\infty, -1]\cup[0, 1]$$
so from above two cases, I get answer: $x\in\mathbb{R}-\{-1<x<0\}$   
Am I correct? If not please spot my error. thank you very much 

If $x\geq 1$, $1-x\leq 0$ while $|x^3-1|\geq 0$ so the inequality is obviously true. Your approach works as well.
The case $x\leq 1$ is well treated, and I agree with your answer.
