Suppose that $\left| a \right| \leq 1$ and $\left| b \right| \leq 1$, is there a nice way, other than a proof by cases, to show that $$\left| \left| a \right|^n - \left| b \right|^n \right| \leq 1?$$ I'm obviously aware of the triangle inequality $\left| \left| a \right| - \left| b \right| \right| \leq \left| a - b \right|$, but this doesn't help me out.
From $|a|^n \le 1$ we get $|a|^n \le 1+|b|^n$, hence $|a|^n-|b|^n \le 1$
In the same way we show: $|b|^n-|a|^n \le 1$
Notice that $|a| \leq 1 \Rightarrow |a|^n \leq 1$ and same for $b$. Now if you subtract two numbers less or equal than one and take the absolute value, it's still less or equal than one.
Correct me if I am wrong but, the inequality you want to prove is equivalent to
$$-1\leq |a|^n-|b|^n\leq 1$$ Now you know that $|a|,|b|\leq 1$ hence $|a|^n,|b|^n\leq 1$. Therefore you have the inequalities $$0\leq |a|^n\leq 1$$ $$0\leq |b|^n\leq 1\Leftrightarrow -1\leq -|b|^n\leq 0$$ Adding the two inequalities yields the result.