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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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Is it asking you to plug in specific values for $ a $ and $ b $ in each case or is it asking you to prove the formula works in each case given the conditions presented in each part? –  anon Jun 9 '11 at 2:49
    
@anon: I think it is asking me to prove that it works given the condition –  confused Jun 9 '11 at 2:50
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1 Answer 1

up vote 3 down vote accepted

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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Oh so that's enough formalism? I think I was confused because it seemed so obvious that I didn't know how to write it down. Thanks! –  confused Jun 9 '11 at 3:16
    
@confused: Maybe if you're happy with my answer, you could mark it as "better" answer too in a couple of days (just to see if someone writes another one). –  a.r. Jun 9 '11 at 17:56
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