Proof of the triangle inequality

  • Thread starter dobedobedo
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  • #1
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Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
Bacle2
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|x+y+z| =|(x+y)+z| ≤ |x+y|+|z| ≤ |x|+|y|+|z| .....
 
  • #3
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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.
 
  • #4
Bacle2
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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|+

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

How does it follow from above that

|x1+x2+...+xn+1|≤|x1|+

|x2|+.....+|xn+1| ?
 
Last edited:

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