Proving Triangle Inequality for $d(x,y)=|x_2-y_2|$ if $x_1=y_1$, $d(x,y)=|x_2|+|y_2|+|x_1-y_1|$ if $x_1$ not equal to $y_1$

Question

\begin{align*} d(x,y)&=|x_2-y_2|&& \text{if }x_1=y_1, \\ d(x,y)&=|x_2|+|y_2|+|x_1-y_1|&& \text{if }x_1\neq y_1 \end{align*}

Show that $d(x,z)\le d(x,y)+d(y,z)$ where $d:R^2*R^2$

Answer 1

Show $d(x,z) \le d(x,y) + d(y,z)$ for the different cases when:

$x_1 = y_1 = z_1\\ x_1 = y_1 \ne z_1\\ x_1 \ne y_1 \ne z_1\\ x_1 = z_1 \ne y_1$

Case 1: $x_1 = y_1 = z_1$

$d(x,z) = |x_2-z_2| = |(x_2 - y_2) + (y_2-z_2)| \le |(x_2 - y_2)| + |(y_2-z_2)| = d(x,y) + d(y,z)$

Case 2: $x_1 = y_1 \ne z_1$

$d(x,y) + d(y,z) = |x_2-y_2| + |y_2| + |z_2| + |y_1-z_1| \ge |x_2|-|y_2| + |y_2| + |z_2| + |y_1-z_1| = |x_2| + |z_2| + |x_1-z_1| = d(x,z)$

I am going to leave the remaining cases for you to prove.