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So, I'm trying to prove by contradiction that if (1) $\lvert a-b \rvert$ < c, then (2) $\lvert a \rvert$ < $\lvert b \rvert$ + c.

I want to show that given (1), $\lvert a \rvert$ = $\lvert b \rvert$ + c cannot be true and nor can $\lvert a \rvert$ > $\lvert b \rvert$ + c, so that (2) must be true.

I started with the former.

$\lvert a \rvert$ = $\lvert b \rvert$ + c$\quad\longrightarrow\quad$ c = $\lvert a \rvert$ - $\lvert b \rvert$ ≥ $\lvert a-b \rvert$$\quad\longrightarrow\quad$$\lvert a-b \rvert$ ≤ c.

Now, does $\lvert a-b \rvert$ ≤ c contradict $\lvert a-b \rvert$ < c? Or is the latter somehow included in the former? Similarly, would proving that $\lvert a \rvert$ ≥ $\lvert b \rvert$ + c cannot be true enough to demonstrate that (2) must be true?

Thanks in advance.

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    Note: $\lvert a\rvert - \lvert b\rvert \leqslant \lvert a-b\rvert$. – 2017-01-31
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    Does $|a−b| ≤ c$ contradict $|a−b| < c$ ? NO. $5 \le 6$ does not contradict $5 < 6$. – 2017-01-31
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    @DanielFischer Shouldn't that be an answer? – 2017-01-31
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    $|a-b|\ge c$ would contradict $|a-b|$|a|\ge |b|+c$ would demonstrate $(2)$. The underlying assumption is that precisely one of $P$ or not $P$ is true, where $P$ is some statement. – 2017-01-31

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No, $|a-b|\le c$ does not contradict $|a-b|

For example, both are true if $a=b=c=1$. When there is a situation where both claims are true, they are by definition not contradictory.

It is also not true in general that $|a|-|b|\ge |a-b|$; consider for example $a=0, b=1$.