How do I show that $ d((x_1, x_2), (y_1, y_2)) = |y_1 - x_1| + |y_2 - x_2|$ is a metric? I've managed to show the first two axioms that are required for a set to be a metric space. But can't seem to get past showing that for $ z_1, z_2 \in \mathbf R$, $$ d((x_1,x_2),(y_1,y_2)) + d((y_1, y_2),(z_1, z_2)) = |y_1 - x_1| + |y_2 - x_2| + |z_1 - y_1| + |z_2 - y_2|$$ but I don't know how to show that this sum is larger than $d((x_1,x_2),(z_1,z_2))$
Sorry if this question is a duplicate.
 A: Hint:
$$d((x_1,y_1),(x_2,y_2)) =|x_1-z_1+z_1-y_1|+|x_2-z_2+z_2-y_2|. $$
A: This is so-called $l^1$ norm, a case of $l^p$ norms defined as $(\Sigma(x_i-y_i)^p)^{1/p}$ for $p>=1$, of which $l^2$ is the familiar Eucledian norm. The inequality you seek was proven for any $p$ by Minkowski: http://en.wikipedia.org/wiki/Minkowski_inequality
The case of $l^1$ norm, however, is much simpler than that of the general $p$. All you have to show is that $|z_1−x_1|\leq |z_1-y_1|+|y_1-x_1|$ because in $l^1$ the norm conveniently splits by coordinates, so you only have to prove it in dimension 1. And 1-D case is obvious for a number of reasons. For once, all $l^p$ coincide in dimension 1, so 1-D inequality for $l^1$ follows from the one for $l^2$. 
Alternatively, you can prove it by considering all the cases of where $y_1$ lies relative to $x_1$ and $z_1$. For example, if $z_1\ge x_1$ and $y_1$ is between them then $|z_1-x_1|=z_1-x_1=(z_1-y_1)+(y_1-x_1)=|z_1-y_1|+|y_1-x_1|$. On the other hand, if $y_1<x_1$ then $|z_1-x_1|=z_1-x_1=(z_1-y_1)+(y_1-x_1)>|z_1-y_1|+|y_1-x_1|$. Etc.
