I'm not very sure how to proceed on this question. I tried breaking the $|a-b|$ into a piecewise function and showing that each of the functions of $|a-b|$ are greater than the right side but I cant figure it out. Any help would be appreciated.
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3Would you find it easier to prove $|a - b| + |b| \ge |a|$? (does this remind you of any identities/inequalities you know?) – Izaak van Dongen Oct 06 '23 at 22:27
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Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Oct 06 '23 at 22:33
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1To @fiends, please ignore the comment from the Bot, your post is fine. – Rob Arthan Oct 06 '23 at 22:36
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Does this answer your question? Triangle inequality for subtraction? - found using an Approach0 search. Note this is closed as a duplicate of another question, and there are other duplicate questions linked to it, e.g., Prove $|a−b| \ge |a|−|b|$. – John Omielan Oct 07 '23 at 00:17
1 Answers
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If you know the triangle inequality
$$|x+y|\le |x|+|y|\tag1$$
then you can simply let $x=a-b$ and $y=b$. Then from $(1)$ we see that
$$|a-b+b|=|a|\le |a-b|+|b|\tag2$$
Rearranging $(2)$ yields
$$|a-b|\ge |a|-|b|$$
and we are done!
Mark Viola
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