1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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,
    Dobedobedo!
     
  2. jcsd
  3. Aug 30, 2012 #2

    Bacle2

    User Avatar
    Science Advisor

    |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

    Bacle2

    User Avatar
    Science Advisor

    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|+

    |x2|+.....+|xn|.

    How does it follow from above that

    |x1+x2+...+xn+1|≤|x1|+

    |x2|+.....+|xn+1| ?
     
    Last edited: Sep 1, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof of the triangle inequality
  1. Triangle inequality (Replies: 12)

  2. Inequality Proof (Replies: 3)

Loading...