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I've been wondering, is there a generalized version of the triangle inequality that is useful in math? I recently saw the definition of a metric space, and wondered what would happen if you want a function that satisfies the first two conditions of a metric function but instead of the third condition, satisfies something like $d(x_{1}, x_{n}) \leq d(x_{1}, x_{2}) + ... + d(x_{n-1}, x_{n}))$? Does an inequality like this show up anywhere else in math? If not, why is it that having this generalized inequality with $n = 3$ so special? I've looked at the polygon example of the wiki article: http://en.wikipedia.org/wiki/Triangle_inequality#Generalization_of_the_inequality_to_any_polygon and tried reading http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_35.pdf but that's above my level right now. Any help/advice is appreciated. Thanks!

Sincerely,

Vien

1 Answers 1

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As the Wikipedia article says, the ordinary triangle inequality implies all of the versions with $n>3$. For example, we can use the ordinary triangle inequality to see that

$\begin{align*} d(x_1,x_4)&\le d(x_1,x_3)+d(x_3,x_4)\\ &\le\Big(d(x_1,x_2)+d(x_2,x_3)\Big)+d(x_3,x_4)\\ &=d(x_1,x_2)+d(x_2,x_3)+d(x_3,x_4)\;, \end{align*}$

and this process can clearly be extended to arbitrarily large $n$. (Technically, this is done by induction.)

Moreover, each of the versions with $n\ge 3$ implies the ordinary version. Again I’ll use $n=4$ as an example. If $d(x_1,x_4)\le d(x_1,x_2)+d(x_2,x_3)+d(x_3,x_4)$ for all $x_1,x_2,x_3,x_4$, then we may in particular take $x_3=x_4$ to get

$\begin{align*} d(x_1,x_4)&\le d(x_1,x_2)+d(x_2,x_3)+d(x_3,x_4)\\ &=\le d(x_1,x_2)+d(x_2,x_3)+d(x_3,x_3)\\ &=\le d(x_1,x_2)+d(x_2,x_3)+0\\ &=\le d(x_1,x_2)+d(x_2,x_3)\;, \end{align*}$

which is the ordinary triangle inequality.

In other words, all of these versions are equivalent, so we might as well stick with the simplest, most easily motivated one for our definition.