Proof of $c > a - 2b$ if $|c - a| < b$

Question

Let $(K,+,\cdot, P)$ be a totally ordered field.

How can one prove that for $a,b,c \in K$ if $|c - a| < b$ it holds that $c > a - 2b$?

I tried it out with numbers and couldn't find a counterexample, so I think it's true. But how can one prove this formally?

Answer 1

Since $0\leq |c-a|<b$ (implying that $b> 0$), we get that: $-b<c-a<b$ implying that $c>a-b$. But since $b>0$ we have that: $$c>a-b> a-2b$$