Motivation for triangle inequality

Solution 1:

Any reasonable notion of distance satisfies the triangle inequality, since if you can get from point $A$ to point $B$ using a route of length $d(A, B)$ and from point $B$ to point $C$ using a route of length $d(B, C)$, you can clearly get from point $A$ to point $C$ using a route of length $d(A, B) + d(B, C)$, and the optimal route (of length $d(A, C)$) therefore can't be any longer than this. (This argument is completely rigorous in an intrinsic metric, but there's no reason not to take it as motivation in general.)

Lawvere noticed that the above argument looks an awful lot like composition of morphisms in a category, and that led him to the categorical definition of metric spaces as certain kinds of enriched categories. From this perspective neither the symmetry nor the positive-definiteness axioms are particularly natural, but the triangle inequality is still very natural.

@Rasmus: letting $x = y$ in the reverse triangle inequality gives $0 \ge 2d(x, z)$, so the only "reverse metric space" is a point.

Solution 2:

It seems I can't add comments so I decided to post this as an answer, I apologize if it seems trivial or redundant. My point of view is that most occurrences of this inequalities are just minimum principles (something you could summarize as "min $\le$ any"). In each case, the quantity (distance, norm, dimension, cardinal...) can be viewed as a minimum :

• The distance between two points is the length of the shortest path between them

• The dimension of a vector space smallest size for a system of generator

• The cardinal of a set is the smallest $n$ for which your set can be embedded in $[\!|1,n|\!]$

...

In that setting, the triangle inequality more or less states that the right hand side correspond to a possibility (of path, generating set, embedding...) :

• A path from $x$ to $z$ and then $z$ to $y$ is a path from $x$ to $y$.

• The union of a generating sets gives a generating set for the sum.

• If you can embed $A$ in $A'$ and $B$ in $B'$ then $A \cup B$ can be embedded in the disjoint union of $A'$ and $B'$

...

Solution 3:

In the context of normed spaces, subadditivity ensures local convexity, and that the Hahn-Banach theorem can be applied. For example, $L^p[0,1]$ with $0<p<1$ is not locally convex, and worse, it has no nonzero continuous linear functionals. If you try to define a "norm" by $\|f\|_p = \left(\int |f|^p\right)^{1/p}$, it is not subadditive, although it does satisfy $\|\lambda f\|_p=|\lambda|\|f\|_p$ if $\lambda$ is scalar. You can instead define a metric by $d(f,g)=\int |f-g|^p$, which gives $L^p$ the structure of an F-space. The triangle inequality is thus fixed, but the scalars no longer come out nice, and local convexity is lost regardless.

There is in fact a "reverse Minkowski inequality": if $f$ and $g$ in $L^p$ are nonnegative, $0<p<1$, then $\|f\|_p+\|g\|_p\leq \|f+g\|_p$.