Generalized triangle inequality

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Homework Help Overview

The discussion revolves around proving the generalized triangle inequality for a sum of numbers, specifically the inequality |x_1 + x_2 + · · · + x_n | ≤ |x_1 | + |x_2 | + · · · + |x_n |. Participants are exploring the application of induction and properties of absolute values in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using mathematical induction as a method to prove the inequality, with some questioning how to extend the proof to cover all cases of n. Others suggest leveraging the convexity of the absolute value function as an alternative approach.

Discussion Status

The discussion is active, with participants sharing different strategies and hints. There is no explicit consensus yet, but various lines of reasoning are being explored, including induction and properties of convex functions.

Contextual Notes

Some participants express uncertainty about the initial steps of their proofs and the assumptions needed for induction. The discussion reflects a focus on foundational principles related to absolute values and inequalities.

CyberShot
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Homework Statement



Show that

|x_1 + x_2 + · · · + x_n | ≤ |x_1 | + |x_2 | + · · · + |x_n |

for any numbers x_1 , x_2 , . . . , x_n

Homework Equations



|x_1 + x_2| ≤ |x_1| + |x_2| (Triangle inequality)

The Attempt at a Solution



I tried using the principle of induction here, but to no avail.

Can I induct on the basis |x_1| ≤ |x_1| ?
 
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Hint: | x_1+x_2 + x_3 | \leq |x_1 + x_2| + | x_3 |
 
so I can write |x_1 + x_2 + x_3| ≤ |x_1| + |x_2| + |x_3|

since |x_1 + x_2| ≤ |x_1| + |x_2|

but how do I cover all the "n" cases?
 
CyberShot said:
so I can write |x_1 + x_2 + x_3| ≤ |x_1| + |x_2| + |x_3|

since |x_1 + x_2| ≤ |x_1| + |x_2|

but how do I cover all the "n" cases?

Use the inductive principle... assume that | \sum_{i=1}^n x_i| \leq \sum_{i=1}^n |x_i| for some some n=k (it is obviously true for n=1, 2 and 3) , and then show that it must then also be true for n=k + 1.
 
CyberShot said:

Homework Statement



Show that

|x_1 + x_2 + · · · + x_n | ≤ |x_1 | + |x_2 | + · · · + |x_n |

for any numbers x_1 , x_2 , . . . , x_n


Homework Equations



|x_1 + x_2| ≤ |x_1| + |x_2| (Triangle inequality)


The Attempt at a Solution



I tried using the principle of induction here, but to no avail.

Can I induct on the basis |x_1| ≤ |x_1| ?

You can use the easily-proven fact that the absolute-value function is convex, in the sense that f(x) satisfies f(\alpha w_1 + (1-\alpha)w_2) \leq \alpha f(w_1) + (1-\alpha) f(w_2) for all \alpha \in [0,1]. Try to prove that
\left| \frac{x_1 + x_2 + \cdots + x_n}{n}\right| \leq \frac{1}{n}|x_1| + \cdots + \frac{1}{n} |x_n|. Hint: induction.

RGV
 

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