Given $A \leq X$ and $B \leq Y$, prove $|A - B| + |X - Y| \leq |A - Y| + |X - B|$

Question

I've been really struggling to prove the following inequality; I'm assuming I'll need to apply the triangle inequality at some point but I can't figure out how to break it down well enough. The inequality is as follows (we can assume we're just using natural numbers)

Given $A \leq X$ and $B \leq Y$, prove $|A - B| + |X - Y| \leq |A - Y| + |X - B|$

Answer 1

Splitting into cases works quite well. Let $S=|A - Y| + |X - B|-(|A - B| + |X - Y|)$ Without loss of generality, we can assume that $A\le B$. There are then $3$ cases:

1. If $A\le X\le B\le Y$, then $S= B-X+Y-A-(Y-X+B-A)=0$.
2. If $A\le B \le X\le Y$, then $S=Y-A+X-B-(Y-X+B-A)=2(X-B)\ge 0$.
3. If $A\le B \le Y \le X$, then $S=Y-A+X-B-(X-Y+B-A)=2(Y-B)\ge 0$.