I am tasked with the following:
Show the following inequality for $\lvert b \rvert < \frac{\lvert a \rvert}{2}$ where $a,b \in \Bbb R$: $$\frac{1}{\lvert a - b \rvert} < \frac{2}{\lvert a \rvert}$$
I'm a bit at a loss as to where to even start with this one. I'm mostly looking for a little nudge.
$|a-b|>\left||a|-|b|\right| \ge |a|-|b| >|a|-|a|/2=|a|/2$. Take the inverse of both sides, and you get what you want.
Subtract $|a|$ from both sides and use $|x| = |x+y-y| < |x+y|+|y|$, so that $|x|-|y| < |x+y|$
WLOG assume $a>0$ then $|b|<\frac{a}{2}$ $$ \frac{1}{|a-b|} = \frac{1}{a-b}<\frac{1}{a-\frac{a}{2}}=\frac 2a=\frac{2}{|a|}.\blacksquare $$
