How to prove the triangle inequality axiom for the metric system $d(x,y) = \lvert\,x - y\,\rvert$

Question

In other words,

How does one prove $d(x,z) \leq d(x,y) + d(y,z)$ for $d(x,y) = \lvert\,x - y\,\rvert$ given $x, y, z \in \mathbb{R}?$

Answer 1

We have $$\lvert\,x-y\,\rvert=\lvert\,x-z+z-y\,\rvert\leq\, \rvert\,x-z\,\rvert+\lvert\,z-y\,\rvert$$ by the triangle inequality.

Answer 2

We have to use triangle inequality $|a+b|\leq |a|+|b|$:

$$d(x,z) = |x-z| = |(x-y)+(y-z)| \leq |x-y|+|y-z| = d(x,y) + d(y,z)$$

Answer 3

Firstly, for $a,b \in \mathbb R$ we have $$|a + b|^2 = (a+b)^2 = a^2 + 2ab + b^2 \leq |a|^2 + 2|a||b|+|b|^2 = (|a|+|b|)^2$$ and by taking the square roots we have $$|a+b| \leq |a|+|b|.$$

Then, by using the above result, for $x,y,z \in \mathbb R$ we have $$|x-y| = |(x-z)+(z-y)| \leq |x-z|+|z-y|.$$ $$\tag*{$Q.E.D.$}$$