[Math] Showing that the Triangle Inequality holds for the $L_\infty$ norm as a metric.

Question

We have a metric on $\mathbb{R}^2$ defined as: $d(x,y) = \max(|x_1-y_1|,|x_2-y_2|)$ where $x = (x_1,x_2)$ and $y = (y_1,y_2)$. To satisfy the triangle inequality, we must show that $\max(|x_1-y_1|,|x_2-y_2|) \leq\max(|x_1-z_1|,|x_2-z_2|) +\max(|z_1-y_1|,|z_2-y_2|)$. I am having trouble formally proving this with the $max$ statements in there without hand waving. We know from properties of the absolute value that $|x+y| \leq |x| + |y|$. Can anyone show me a formal way to do this?

Answer 1

Cases:

1. $|x_1-z_1|<|x_2-z_2|$ and $|z_1-y_1|<|z_2-y_2|$ (or both are reversed): then by your assumed triangle inequality $|x_1-y_1|<|x_2-y_2|$ (resp. reversed) and the rest is easy.
2. the remaining case, $|x_1-z_1|<|x_2-z_2|$ but $|z_1-y_1|>|z_2-y_2|$ (or if both are reversed, this goes through the same way m.m.): Consider $\max(|x1-y1|,|x2-y2|)$, WLOG $|x2-y2|$. By your assumed triangle inequality it's $<|x_2-z_2|+|z_2-y_2|$, but that's $<|x_2-z_2|+|z_1-y_1|=\max(|x_1-z_1|,|x_2-z_2|)+\max(|z_1-y_1|,|z_2-y_2|)$ qed.