Prove |a|<2|a-b| if 2|b|<|a|

Question

I have already proved the triangle inequality

$\vert a+b \vert \le \vert a \vert + \vert b \vert$

I also proved that

$\vert a \vert - \vert b \vert \le \vert a-b \vert $

and that

$\vert \vert a \vert - \vert b \vert \vert \le \vert a-b \vert $

Now I just need help proving that

if $ 2\vert b \vert \lt \vert a \vert $ then $\vert a \vert \lt 2\vert a-b \vert$

Answer 1

$$2|a-b|\geq2||a|-|b||=2||a|-|b||-|a|+|a|=$$ $$=\frac{4(|a|-|b|)^2-|a|^2}{2||a|-|b||+|a|}+|a|=\frac{(|a|-2|b|)(3|a|-2|b|)}{2||a|-|b||+|a|}+|a|>|a|.$$

Answer 2

Using the "reverse triangle equality" (the second one in your list) you get $$ 2|a -b | \ge 2 |a| - 2 |b| > 2 |a| - |a| = |a| $$