Proof of the triangle inequality

[dobedobedo]

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!

 

[Bacle2]

|x+y+z| =|(x+y)+z| ≤ |x+y|+|z| ≤ |x|+|y|+|z| ...

 

[dobedobedo]

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.

 

[Bacle2]

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+...+x_n|≤|x₁|+

|x₂|+...+|x_n|.

How does it follow from above that

|x₁+x₂+...+x_n+1|≤|x₁|+

|x₂|+...+|x_n+1| ?

 

[CellCoree]

Dear Dobedobedo,

Thank you for your question. The proof for the triangle inequality for an arbitrary number of terms can be done using mathematical induction. Here is the step by step solution:

Step 1: Base case - n=2
As you mentioned, the Cauchy-Schwartz inequality can be used to prove the triangle inequality for two terms. So for n=2, we have:

|x_1+x_2| ≤ |x_1|+|x_2|

Step 2: Assume the inequality is true for n=k
Next, we assume that the inequality is true for n=k, meaning that:

|x_1+x_2+...+x_k| ≤ |x_1|+|x_2|+...+|x_k|

Step 3: Prove it for n=k+1
Now, we need to prove that the inequality holds for n=k+1. This can be done by using the triangle inequality for two terms (n=2) and the assumption from step 2:

|x_1+x_2+...+x_k+x_{k+1}| = |(x_1+x_2+...+x_k)+x_{k+1}| ≤ |x_1+x_2+...+x_k|+|x_{k+1}| ≤ |x_1|+|x_2|+...+|x_k|+|x_{k+1}|

This shows that the inequality holds for n=k+1, and thus completes the proof by mathematical induction.

Step 4: Conclusion
Therefore, we can conclude that the triangle inequality holds for any number of terms, as it holds for n=2 and we have shown that it holds for n=k+1 assuming it holds for n=k. This completes the proof.

I hope this helps to clarify the process of proving the triangle inequality for an arbitrary number of terms. Please let me know if you have any further questions.

Best regards,