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I am trying to show that $|a+b|\geq||a|-|b||$.

I considered doing an exhaustive process, that is, $a+b \geq 0$, $a+b \leq 0$, and $a+b=0$, and then for each of those considering $|a|-|b| \geq 0$, $|a|-|b| \leq 0$, and $|a|-|b|=0$ (9 combinations)

After doing the first of these ($a+b \geq 0$, $|a|-|b| \geq 0$), I came to the following: $$a+b \geq |a|-|b|$$ and attempting to show this was true. However, I am unable to show that this is the case.

Additionally, going through such an exhaustive process simply strikes me as wrong or unnecessarily cumbersome. Any suggestions are appreciated.

  • You can probably do it by separating in cases, however you will have about $8$ of them or so (depending on the sign of $a$, $b$ and $|a|-|b|$). A faster way is the following: Show that $|a|-|b|\leq |a-b|$ (this is very easy). By symmetry, we can swap $a$ and $b$, and obtain $|b|-|a|\leq |a-b|$. Therefore $-|a-b|\leq |a|-|b|\leq |a-b|$, which is equivalent to $||a|-|b||\leq|a-b|$. – Luiz Cordeiro Sep 17 '16 at 20:53

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An easier approach is to start with $$ |a| = |a+b-b| \leq |a+b|+|b| \Rightarrow |a|-|b| \leq |a+b|$$ and then interchange $a$ and $b$ to get $|b|-|a| \leq |a+b|$ from which the inequality follows.

H. H. Rugh
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  • Thank you. I had the triangle inequality at my disposal but was unable to figure out how to use it. Do you have any tips on how to think about these problems in a better way? I always seem to try to go about solving them in ridiculous ways which end up taking an eternity (as above). –  Sep 17 '16 at 20:55
  • Not really. The above is a standard trick used for any normed vector space, but I don't think it is so obvious to think of the trick in the first place, mainly because of the abs value of $|a|-|b|$ which prevents you from getting the idea of moving things around. – H. H. Rugh Sep 17 '16 at 21:00