Triangle inequality. Where does it come from in this form?

Question

I always understood the triangle inequality to be the following:

If you have 2 sides of a triangle, say 4 and 6, the third side had to be between 2 and 10.

However, I see the triangle inequality written in this form:

$|a+b| \leq |a|+ |b| $

How do you prove it?

I know that:

$-|a| \leq a \leq |a|$

and

$-|b| \leq b \leq |b|$

and adding you get:

$-(|a| + |b|) \leq a + b \leq |a| + |b|$

Now what?

I am learning this because I want to learn the Sum Law proof:

$\lim_{x \to a} [f(x) + g(x)] = L + M$

The first step is showing that

$|f(x) + g(x) - (L + M)| \leq |f(x) - L| + |g(x) - M|$ using epsilon delta

Answer 1

$$-(|a| + |b|) \leq a + b \leq |a| + |b|$$ implies $$|a+b|\le |a|+|b|$$

Answer 2

Consider that $$ |x+y| \leq |x|+|y| \Leftrightarrow |x+y|^2 \leq (|x|+|y|)^2 $$ $$\Leftrightarrow x^2+2xy+y^2 \leq x^2+2|x|.|y|+y^2 $$ $$\Leftrightarrow xy \leq |x|.|y| $$ $$\Leftrightarrow xy \leq |xy| $$

The last inequality is always true.

Answer 3

Take $|a+b|\leq |a|+|b|$ and replace "$a$" with "$x-y$" and "$b$" with "$y-z$":

$|x-z| \leq |x-y|+|y-z|$

Interpret something like $|x-y|$ as the distance between $x$ and $y$, where $x,y$ do not have to be real numbers, and $|\cdot|$ does not have to be the standard Euclidean metric.

This is the general idea of the triangle inequality, and it now applies to your standard example of side lengths. Interpret $x,y,z$ as the points of a triangle on the plane, and so e.g. $|x-y|$ is the length of the side between the points $x$ and $y$. The statement now says, "the combined length of two sides of a triangle must be longer than the 3rd side".

I also prefer to share this form of the inequality when presenting it to students, as well, because there is a useful, colloquial heuristic for remembering it: Stopping by Starbucks on your way to class cannot be a shortcut.