# Proof Review for Triangle Inequality for $d((x_1, x_2), (y_1, y_2)) = max( |y_1 - x_1|, |y_2 - x_2|)$

Let $X = \mathbb{R}^2$ and define a function $d : \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ by $d((x_1, x_2), (y_1, y_2)) = \max( |y_1 - x_1|, |y_2 - x_2|)$, where $\max(a, b)$ is the maximum of $a$ and $b$.

I can't tell if the following proof is correct and would appreciate it if people could please take the time to review it.

At each step, I have done my best to explain my reasoning. If there is anything about my understanding that is incorrect and/or lacking, I would be grateful for explanation.

The Proof

Let $x = (x_1, x_2), y = (y_1, y_2), z = (z_1, z_2)$.

We want to show that $d(x, z) \le d(x, y) + d(y, z)$; in other words, that the function $d$ satisfies the triangle inequality.

$d(x, z) = \max(|z_1 - x_1|, |z_2 - x_2|)$ (By the definition of $d$.)

We will begin by looking at $|z_1 - x_1|$ and $|z_2 - x_2|$ separately.

1. The case for $|z_1 - x_1|$ follows.

$|z_1 - x_1| \leq |z_1 - y_1| + |y_1 - x_1|$ (By the triangle inequality for $\mathbb{R}$.)

$\leq \underbrace{\max(|z_1 - y_1|,|z_2 - y_2|)}_\text{This is$d(y, z)$, which is$\ge \ |z_1 - y_1|$} \ + \underbrace{\max(|y_1 - x_1|,|y_2 - x_2|)}_\text{This is$d(x, y)$, which is$\ge \ |y_1 - x_1|$}$

1. The case for $|z_2 - x_2|$ follows.

$|z_2 - x_2| \leq |z_2 - y_2| + |y_2 - x_2|$ (By the triangle inequality for $\mathbb{R}$.)

$\leq \underbrace{\max(|z_1 - y_1|, |z_2 - y_2|)}_\text{This is$d(y, z)$, which is$\ge \ |z_2 - y_2|$} \ + \underbrace{\max(|y_1 - x_1|, |y_2 - x_2|)}_\text{This is$d(x, y)$, which is$\ge \ |y_2 - x_2|$}$

Now we bring it all together to state the conclusion:

$\therefore d(x, z) = \max(|z_1 - x_1|, |z_2 - x_2|)$

$\leq \underbrace{\max(|y_1 - x_1|, |y_2 - x_2|)}_\text{This is$d(x, y)$} \ + \underbrace{\max(|z_1 - y_1|,|z_2 - y_2|)}_\text{This is$d(y, z)$} = d(x,y) + d(y, z)$

I've spent a tremendous amount of time on trying to understand this problem, so I would greatly appreciate reviews.

Since the kind reviewers are indicating that the proof is correct, I want to write a little note-to-self (and future viewers) that explains what I was misunderstanding.

What Was My Misunderstanding?

What I wasn't seeing was that we needed to show that each term in $d(x, z) = \max(|z_1 - x_1|, |z_2 - x_2|)$ -- $|z_1 - x_1|$ and $|z_2 - x_2|$ -- must be $\le d(x, y) + d(y, z)$. Why? Because if we show that $|z_1 - x_1| \le d(x, y) + d(y, z)$ and $|z_2 - x_2| \le d(x, y) + d(y, z)$, then we have that $d(x, z) = \max(|z_1 - x_1| \le d(x, y) + d(y, z), |z_2 - x_2| \le d(x, y) + d(y, z))$. By the properties of the $\max()$ function, this implies that $d(x, z)$ can at most be $\le d(x, y) + d(y, z)$! So essentially what we've done here is found an upper bound for each of the arguments of the $\max()$ function, $|z_1 - x_1|$ and $|z_2 - x_2|$.

• Looks very precise to me, and correct. Good job :) – 57Jimmy Apr 17 '18 at 14:12
• Your calculation looks precise and the have done a great job. – WKhan Apr 17 '18 at 14:30
• Thank you all for the review and kind words. – The Pointer Apr 17 '18 at 14:30
• +1 for the self note. – user370967 Apr 19 '18 at 9:38

For real $e,f,g,h$ we have $$(i)...\;\quad e+f\leq e+\max (f,h)\leq\max (e,g)+\max (f,h).$$ $$(ii)...\;\quad g+h\leq g+\max (f,h)\leq \max (e,g)+\max (f,h).$$ Therefore $\max (e+f,g+h)\leq \max (e,g)+\max(f,h).$
With $e=|x_1-y_1|$ and $f=|y_1-z_1|$ and $g=|x_2-y_2|$ and $h=|y_2-z_2|$ we have $|x_1-z_1|\leq e+f$ and $|x_2-z_2|\leq g+h.$ Therefore $$d((x_1,x_2),(z_1,z_2))=\max (|x_1-z_1|,|x_2-z_2|)\leq$$ $$\leq \max (e+f,g+h)\leq$$ $$\leq \max (e,g)+\max (f,h)=$$ $$=d((x_1,x_2),(y_1,y_2))+d((y_1,y_2),(z_1,z_2)).$$