Generalized triangle inequality

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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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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 [itex]| \sum_{i=1}^n x_i| \leq \sum_{i=1}^n |x_i|[/itex] for some some [itex]n=k[/itex] (it is obviously true for n=1, 2 and 3) , and then show that it must then also be true for [itex]n=k + 1[/itex].
 
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 [itex]f(\alpha w_1 + (1-\alpha)w_2) \leq \alpha f(w_1) + (1-\alpha) f(w_2)[/itex] for all [itex]\alpha \in [0,1].[/itex] Try to prove that
[tex]\left| \frac{x_1 + x_2 + \cdots + x_n}{n}\right| \leq \frac{1}{n}|x_1| + \cdots + \frac{1}{n} |x_n|.[/tex] Hint: induction.

RGV