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?
On
$|z^2-1| \ge |z+|1-z^2||$
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.
On
Note that for any $a$ and $b$, we have $|a|\ge |b|$ iff $a^2 \ge b^2$. Apply this with $a$ the left-hand side, and $b$ the right-hand side of our expression. Thus our inequality is equivalent to $$(z^2-1)^2\ge z^2+2z|1-z^2| +(1-z^2)^2.$$ Since $(z^2-1)^2=(1-z^2)^2$, we are trying to solve the inequality $$z^2+2z|1-z^2| \le 0.\tag{$1$}$$ Sure killed an awful lot of absolute value signs!
The inequality $(1)$ holds at $z=0$. And it is obvious that it cannot hold for positive $z$. So (remembering that from now on $z$ is negative), we are looking at the inequality $$z+2|1-z^2| \ge 0.$$ The rest is routine. We can divide into two cases, $z\le -1$ and $-1\lt z\lt 0$. It turns out that the inequality holds for all $z \le 0$, except for the numbers in the open interval $(a,b)$, where $a=-\frac{\sqrt{17}+1}{4}$ and $b= -\frac{\sqrt{17}-1}{4}$.
On
Here is a solution
$$|z+|1-z^2||\leq |1-z^2| \implies -|1-z^2|\leq z+|1-z^2|\leq |1-z^2| $$
$$ \implies -2|1-z^2|\leq z\leq 0 \,. $$
From the above inequality, the solution $z$ should lie in the set $ \left\{z\leq 0\right\} \cap \left\{ z\geq-2|1-z^2| \right\} $.
working out $ z \geq -2|1-z^2|, $ gives
$$\left\{z \geq -2|1-z^2| \right\} = \left\{z \geq -2(1-z^2) \right\} \cup \left\{z \geq -2(-1+z^2) \right\}$$
$$ = ( -.78, 1.28 ) \cup \left\{ (-\infty, -1.28)\cup (-.78,\infty) \right\} $$
$$ = (-\infty, -1.28) \cup ( -.78, \infty ) . $$
Thus, the solution is given by
$$ \left\{z\leq 0\right\} \cap \left\{ z\geq -2|1-z^2| \right\}$$
$$=\left\{z\leq 0\right\} \cap \left\{ (-\infty, -1.28) \cup ( -.78, \infty ) \right\} $$
$$ =\left\{\left\{z\leq 0\right\} \cap (-\infty, -1.28)\right\} \cup \left\{\left\{z\leq 0\right\} \cap ( -.78, \infty )\right\} $$
$$ = \left( -\infty,-1.28\right) \cup \left(-.78, 0 \right).$$
Note: I approximated the roots when I was solving the inequalities.
I am presuming that $z$ is real. The problem is that the outer absolute on the right may change sense at other places. Say $z \lt -1$. Then $|z+|1-z^2||=|z+z^2-1|$, but now you are testing whether $z+z^2-1 \gt 0$ which doesn't change sense at those points. So you need to find some secondary cases based on what you get for the prime cases.