The task in the following problem is to prove the implication using the Triangle Inequality:
Show that $|b|<\frac{1}{2} \implies |b^3-1|>\frac{7}{8}$.
I tried the following:
- By the Triangle Inequality:
$$ |b^3-1| \ge ||b^3|-|1|| = \color{blue}{ ||b^3|-1| \ge |b|^3-1 } $$
- Since $|b|<\frac{1}{2} \implies |b|^3<\frac{1}{8} \implies |b|^3 - 1<-\frac{7}{8}$
But that's not correct. I went wrong at blue, highlighted part, in the solution provided in my book they do it differently:
$$ |b^3-1| \ge ||b^3|-|1|| = \color{blue}{ ||b^3|-1| \ge 1-|b|^3 } $$
I understand that when proofing such inequalities, it's good to simplify as much as possible to make the variable, here $|b|$, stand outside of absolute values. But I can't follow here why this simplification was allowed.
Is this based on some rule how to resolve absolute values? I can't seem to get the reasoning why they could do $... \ge 1-|b|^3$ instead of $... \ge |b|^3-1$ what I tried.