Exercise 1.2.11

[muzimuzhi]

Exercise 1.2.11

[muzimuzhi]

Solution: By the triangle inequality, we know
$$\begin{align*}
\lvert x-y \rvert &\le \lvert x-a \rvert + \lvert a-y \rvert, \\
\lvert x-y \rvert &\le \lvert x-b \rvert + \lvert b-y \rvert, \\
\lvert a-y \rvert &\le \lvert a-c \rvert + \lvert c-y \rvert, \text{ and} \\
\lvert b-y \rvert &\le \lvert b-d \rvert + \lvert d-y \rvert.
\end{align*}$$
Add them up, and notice that
$$\begin{align*}
\lvert x - a \rvert + \lvert x - b \rvert &= b - a \text{ and} \\
\lvert y - c \rvert + \lvert y - d \rvert &= d - c,
\end{align*}$$
we directly have
$$
2\lvert x-y \rvert = (b-a) + (d-c) + \lvert c-a \rvert + \lvert d-b \rvert.
$$

inspired by https://math.stackexchange.com/q/1789702