51st Ukrainian National Mathematical Olympiad, 4th Round

It is clear that $$\max \{|x|-1,x^2-1\}\ge x^2-1\ge \min \{x^2-1,1-x\},$$ 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 $$|x|-1\le x^2-1\le 1-|x|.$$ Therefore, $1-|x|\ge |x|-1$, which means that $|x|\le 1$. Also we should have that $|x|\le |x{|}^2=|x|\cdot |x|$, and since we already know that $|x|\le 1$, 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 $x$.