<s>[INST] double absolute values I am having a little bit of problem with an inequality with nested absolute values:
$$|z^2-1| \ge |z+|1-z^2||$$
I've tried solving it by making three cases, $z\ge1$, $z\le-1$ and $z$ between $1$ and $-1$ and thus getting rid of absolute values for $z^2-1$ and $1-z^1$, and I am only left with 1 absolute value. But solutions at the end are not what they should be based on the graph.  Here, $z$ is real, and WolframAlpha gives this solution.
What I am doing wrong?
 [/INST] $|z^2-1| \ge |z+|1-z^2||$
Case 1: Suppose $z \ge 1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
Also $z + (z^2 - 1) > 0$ so:
$z^2-1 \ge z+(z^2-1)$
$0 \ge z$
This is a contradiction.
Case 2: Suppose $z \le -1$. Then $|z^2 - 1| = z^2 - 1$ and $|1 - z^2| = z^2 - 1$:
$z^2-1 \ge |z+(z^2-1)||$
There is a root of $z^2 + z - 1$, so we must case on that.
Case 2a: Suppose $z \le -\frac{\sqrt5 + 1}{2}$, then:
$z^2-1 \ge z^2+z-1$
Also a contradiction.
Case 2b: Suppose $-\frac{\sqrt5 + 1}{2} \le z \le -1$, then
$z^2-1 \ge -z^2-z+1 \Rightarrow 2z^2 + z \ge 0$. $z \le -\frac{\sqrt5 + 1}{2}$ always satisfies this.
Case 3: Suppose $-1 \le z \le 1$, then
$1-z^2 \ge |z+1-z^2)|$
This has a root at $\frac{1-\sqrt5}{2}$, so we case there,
Case 3a: $\frac{1-\sqrt5}{2} \le z \le 1$
$1-z^2 \ge z+1-z^2)$
$0 \ge z$. $\frac{1-\sqrt5}{2} \le z \le 0$ satisfies this.
Case 3b: $-1 \le z \le \frac{1-\sqrt5}{2}$.
$1-z^2 \ge z^2-z-1$
$2z^2 - z \le 0$. This does not hold for negative $z$, so it is a contradiction.
We conclude that $z \le -\frac{\sqrt5 + 1}{2}$ or $\frac{1-\sqrt5}{2} \le z \le 0$.
The graphing method is definitely easier here. It also may be easier to consider the potential roots first and then use more cases instead of cases-with-subcases, though ultimately those are similar arguments.
</s>