How to solve $|x^2-1|-2\ge 2x$ I am trying to solve the  inequality
$$|x^2-1|-2\ge 2x$$
but I am not sure where to start, because I have $x²$ in the absolute value part. $x²$ is always positive, and this is confusing for me. Can you please explain how to solve it?
 A: Another way this can be written:
One case is  
$$ (x^2 - 1) \ - \ 2  \ \ge \ 2x \ \ \Rightarrow \ \  x^2 \ - \ 2x \ - \ 3 \ \ge \ 0 \ \ \Rightarrow \ \ (x + 1) \ (x - 3 ) \ \ge \ 0 $$
and the other is
$$ (1 - x^2) \ - \ 2  \ \ge \ 2x \ \ \Rightarrow \ \  0 \ \ge \ x^2 \ + \ 2x \ + \ 1 \  \ \Rightarrow \ \ 0 \ \ge \ (x + 1)^2 \ \ . $$
A: Factor $x^2 - 1$ and add $2$ on both sides to obtain
$|(x+1)(x-1)| \geq 2(x+1)$.
Now there are two cases. First, if $x \leq -1$, the RHS $2(x+1)$ is non-positive and hence the inequality is definitely fulfilled.
Second, let $x > -1$. But then, it follows that $|(x+1)(x-1)| = (x+1)|x-1|$, since $(x+1)$ is positive.
The inequality thus boils down to $(x+1)|x-1| \geq 2(x+1)$. 
Eliminating the positive (by assumption of $x > -1$) factor $(x+1)$ then yields
$|x-1| \geq 2$. 
This is either the case if $x-1 \geq 2$ ($\implies x \geq 3$), or $x-1 \leq -2$ ($\implies x\leq -1$). The second contradicts our assumption of $x > -1$, so $x\geq 3$ remains.
We conclude that the inequality is fulfilled if either (a) $x \leq -1$ or (b) $x \geq 3$.
A: $$|x^2 - 1| -2\ge 2x \iff |x^2 - 1| \ge 2x + 2 = 2(x + 1)$$ $$\iff |(x - 1)(x + 1)|\ge 2(x+1)$$
Now consider the cases $x^2\geq 1$ and $x^2 \lt 1$.
