Prove $\max \{|x_1 - y_1|,\dots,|x_n - y_n|\}$ is a metric space in $\mathbb{R}^n$

Question

I would greatly appreciate someone lending a hint as to how to start the proof for verifying the 4th condition for a function $d(x,y)$ to be a metric (triangle inequality). I tried starting by saying, "fix points $x$ and $y$, where $x = (x_1,...,x_n)$ and $y = (y_1,...,y_n)$ and assume that the $max = |x_k - y_k|$". I don't know where to go from here. I was think about using a case by case proof. For instance, let $z = (z_1,...,z_n)$ and assume $|x_r - z_r|$ is the max for $d(x,z)$ and $|z_t - y_t|$ is the max for $d(z,y)$ and comparing what happen, say, if $x_k < x_r < z_r < z_t < y_t$, but that seems much to complicated and overkill.

Answer 1

Triangle inequality says that

$$d(x,y) \le d(x,z) + d(z,y)$$

That is you want to prove that

$$\max_i |x_i-y_i|\le \max_i |x_i-z_i| + \max_i|z_i-y_i|$$

Suppose the index that attains the maximum for the LHS is $j$.

Clearly by triangle inequality, we have $$|x_j - y_j | \le |x_j - z_j | + |z_j - y_j |$$

Now observe that we have $$|x_j - z_j | \le \max_i |x_i-z_i|$$

and similarly for the other term.

Answer 2

So here is a hint. We want to prove that for all $x$, $y$, and $z$, $$d(x,z) \leq d(x,y) + d(y,z).$$

By definition, $d(x,z) := \max_i |x_i-z_i|.$ So $d(x,z) = |x_i-z_i|$ for some integer $i$ with $1\leq i\leq n$.

Notice that $|x_i-z_i| = |x_i-y_i + y_i-z_i| \leq |x_i-y_i| + |y_i-z_i|.$

I think that will help you.