Prove: For any real number $a$, $b$ : $|a - b| \ge |a| -|b|$

Question

I think this question has to do with field axioms so was wondering if you can claim that $$|a - b| - |a| - |b| \ge 0$$ and go from there.

Answer 1

It follows from the triangle inequality if I'm not wrong:

$|a-b| = |a + (-b)| \geq |a| + |-b| \geq |a| - |b|$

Answer 2

By the triangle inequality we have

$$|a|=|(a-b)+b|\le |a-b|+|b|$$ so $$|a|-|b|\le |a-b|$$

Answer 3

One can use the triangle inequality of course, but I think it is more instructive to instead use the fact that, given fixed magnitudes for $a,b$, the quantity $|a - b|$ is maximized when $a,b$ have opposite signs and minimized when they have the same sign. You can then work through all the cases.

Answer 4

Note that $ |a| = |(a-b)+b| \le |a-b| + |b| $, and therefore one have that $|a|-|b|\le|a-b|$.