# How does the triangle inequality correspond to a triangle?

In the very first chapter in my class "mathematical analysis 1" I've seen something called the triangle inequality, which is $$||a| - |b|| \leq |a \pm b| \leq |a| + |b|$$. Now the thing is that I do understand why this is true, but i fail to see what this actually has to do with triangles. It was never explained nor proven. I'd like to know how this inequality relates to triangles to get a better understanding of it, because it's also used a lot further on. Would someone be able to explain this?

• The name makes more sense in the more general case where you replace scalars with vectors and absolute value with euclidean norm (or any other norm). Dec 24, 2018 at 12:32
• A plot like this might be useful: upload.wikimedia.org/wikipedia/commons/0/08/… (NB: this is for the case where $a,b$ are vectors and $|a|$ denote the length of the vector $a$) Dec 24, 2018 at 12:34

In Euclidean space with Cartesian coordinates, let $$+$$ denote vector addition, and let $$O$$ denote the origin. Choose points $$a,b$$, and form a triangle with vertices $$a,b,O$$. One side of this triangle has side length $$|a|$$, another has side length $$|b|$$, and the third has side length $$|a-b|$$. Now as we all know, the shortest path between two points is a straight line. So, the shortest path between the two points $$a,b$$ has length $$|a-b|$$. There is a longer path one can take which goes around the other two sides, the length of that path is $$|a| + |b|$$. So, $$|a-b| \le |a| + |b|$$.
Suppose $$a$$ and $$b$$ are complex numbers, so that they correspond in the Argand-Cauchy plane, to two sides of a triangle, the third side corresponding to $$a-b$$. Then these inequalities correspond to well-known from middle school inequalities about triangles – namely that the length of a side of a triangle is between the difference and the sum of the lengths of the two other sides.