It is clear that
$$\max\{|x|-1, x^2-1\} \geq x^2-1 \geq \min\{x^2-1, 1-x\},$$
so the inequality from the problem condition can be satisfied only if
$$\max\{|x|-1, x^2-1\} = x^2-1 = \min\{x^2-1, 1-x\},$$
and this, in turn, implies that
$$|x|-1 \leq x^2-1 \leq 1-|x|.$$
Therefore, $1-|x| \geq |x|-1$, which means that $|x| \leq 1$. Also we should have that $|x| \leq |x|^2 = |x| \cdot |x|$, and since we already know that $|x| \leq 1$, this can happen only if $|x|=1$ or $|x|=0$, so $x=1, x=0$ or $x=-1$.

It is also easy to solve this problem using graphs (fig. 35). Here the graph of the left-hand side is drawn with dash-and-dash line, and the graph of the right-hand side – with dash-and-dot line. They intersect exactly at the specified values of $x$.