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Proof of the triangle inequality

  1. Aug 30, 2012 #1
    I am familiar with the proof for the following variant of the triangle inequality:

    |x+y| ≤ |x|+|y|

    However, I do not understand the process of proving that there is an equivalent inequality for an arbitrary number of terms, in the following fashion:

    |x_1+x_2+...+x_n| ≤ |x_1|+|x_2|+...+|x_n|

    How do I prove this? Please write down the solution step by step. I know that the Cauchy-Schwartz inequality is used for proving this in the case where n=2, but is it possible to use it for any n?

    Thank you,
  2. jcsd
  3. Aug 30, 2012 #2


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    |x+y+z| =|(x+y)+z| ≤ |x+y|+|z| ≤ |x|+|y|+|z| .....
  4. Sep 1, 2012 #3
    Haha okay. I get the way of solving it thanks, but that answer is just lazy haha. I guess induction should be use somehow? Okay. I'll try to figure it out.
  5. Sep 1, 2012 #4


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    Sorry if it is not too formal; I thought I'd give you the idea. Well...., yes, you got me,

    I was being lazy too.

    But, yes,you could do an induction on the number of terms:

    Assume |x1+x2+...+xn|≤|x1|+


    How does it follow from above that


    |x2|+.....+|xn+1| ?
    Last edited: Sep 1, 2012
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