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I am going through some practice problems in Abbott's Analysis text, and one of them is the following: Verify the triangle inequality in the special case where

a) a and b have the same sign;
b) $a \geq 0$, $b < 0$, and $a+b \geq 0$.

I do not know where to begin because I don't know how exactly to go about the 'verification' given such conditions. Thanks!

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    @anon: I think it is asking me to prove that it works given the condition2011-06-09

1 Answers 1

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For instance, in case (a), you could distinguish between two possibilities:

  1. $a,b \geq 0$. Then $a+b \geq 0 $ too and $\mid a+b\mid = a+b$ which is certainly equal to $\mid a \mid +\mid b\mid = a +b$.
  2. $a,b \leq 0$. Then $a+b \leq 0$ too and $\mid a+b\mid = -(a+b)$ which is certainly equal to $\mid a \mid +\mid b\mid = -a -b$.
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    @co$n$fused: Maybe if you're happy with my answer, you could mar$k$ it as "better" answer too in a couple of days (just to see if someone writes another one).2011-06-09