Why $|A_1 - A_2| + |B_1 - B_2|$ would not be greater than both $|A_1 - B_1| + |A_2 - B_2|$ and $|B_1 - A_2| + |A_1 - B_2|$ at the same time?

Question

Here is an example:

$A_1, A_2, B_1, B_2 = 5, 8, 0, 9$ respectively

$|A_1 - A_2| + |B_1 - B_2| = 12$

$|A_1 - B_1| + |A_2 - B_2| = 6$

$|B_1 - A_2| + |A_1 - B_2| = 12$

For any $A_1, A_2, B_1, B_2 \in \mathbb{R}$, $|A_1 - A_2| + |B_1 - B_2|$ would not be greater than both of the other two sums. How can this be proven?

Answer 1

Let $(a,b,c,d)$ be a permutation of $(A_1,A_2,B_1,B_2)$ such that $a \leq b \leq c \leq d$. Then each of the sums \begin{align} X&=|A_1-A_2|+|B_1-B_2|, \\ Y&=|A_1-B_1|+|A_2-B_2|, \\ Z&=|B_1-A_2|+|A_1-B_2|, \end{align} corresponds to one of the sums \begin{align} x&=(b-a)+(d-c), \\ y&=(c-a)+(d-b), \\ z&=(d-a)+(c-b). \end{align} Clearly, $x=\min\{x,y,z\}$. In addition, \begin{align} y&=((c-b)+(b-a))+((d-c)+(c-b))=(b-a)+2(c-b)+(d-c), \\ z&=((d-c)+(c-b)+(b-a))+(c-b)=(b-a)+2(c-b)+(d-c), \end{align} so $y=z$. It follows, therefore, that the largest two sums among $X,Y,Z$ are equal, so none of them is greater than the other two.

Answer 2

Let $x := A_1 - A_2, y := B_1 - B_2$.

By triangle inequality, we have $$|A_1 - B_1| + |A_2 - B_2| \ge |A_1 - B_1 - (A_2 - B_2)| = |x - y|$$ and $$|B_1 - A_2| + |A_1 - B_2| \ge |B_1 - A_2 + A_1 - B_2| = |x + y|.$$

If $xy \ge 0$, then $|x + y| = |x| + |y|$. If $xy < 0$, then $|x - y| = |x| + |y|$.

Thus, it is impossible that $|A_1 - A_2| + |B_1 - B_2| > |A_1 - B_1| + |A_2 - B_2|$ and $|A_1 - A_2| + |B_1 - B_2| > |B_1 - A_2| + |A_1 - B_2|$ at the same time.