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?
Best Answer
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 $b= -\frac{\sqrt{17}-1}{4}$.