Prove $a

Question

Given: $a<b,c<d$
To Prove: then $|a-c|+|b-d|<|a-d|+|b-c|$
Proof(Attempt): $$|a-c|+|b-d|=|a-c+d-d|+|b-c+c-d|\\\le |(a-d)+(c-d)|+|(b-c)+(c-d)|\\ \stackrel{?}\le|a-d| +|b-c|+\underbrace{(c-d)+(d-c)}_{?} $$

Answer 1

We can assume $a=0$ (otherwise reduce all numbers for $a$)

$$|c|+|b-d|<|d|+|b-c| \;\;\;/ ^2$$ $$c^2+b^2+d^2-2bd+2|c||b-d| < d^2+b^2+c^2-2bc +2|d||b-c|$$ $$-bd+|c||b-d| < -bc +|d||b-c|$$ $$|c||b-d|-|d||b-c| < b(d-c)$$

Last one is true since we have, by triangle inequality ($|x|-|y|\leq|x-y|$): $$|c||b-d|-|d||b-c| \leq |bc-dc -db+cd| = b|c-d| $$

Answer 2

You can only show that $$ \tag{*} |a-c|+|b-d| \le |a-d|+|b-c| $$ and equality can hold, e.g. for $(a, b, c, d) = (1, 2, 3, 4)$. More precisely, we have

Let $a < b$, $c < d$ be real numbers, and denote by $I, J$ the open intervals $I = (a, b)$ and $J = (c, d)$. Then $$ \tag{**} |a-c|+|b-d| + 2 m(I \cap J) = |a-d|+|b-c| $$ where $m(I \cap J)$ is the length of the intersection of the intervals, or zero if intervals are disjoint.

It follows that $(*)$ holds, with equality exactly if the intervals $I, J$ are disjoint.

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For the proof of $(**)$ consider two cases where $I$ and $J$ are disjoint, and a third case where they do intersect:

Case 1: $b \le c$. Then $m(I \cap J) = 0$ and $$ |a-c|+|b-d| = (c-a) + (d-b) = (d-a) + (c-b) = |a-d|+|b-c| \, . $$

Case 2: $d \le a$. Again $m(I \cap J) = 0$, and $$ |a-c|+|b-d| = (a-c) + (b-d) = (a-d) + (b-c) = |a-d|+|b-c| \, . $$

Case 3: $b > c$ and $d > a$. Then $I \cap J$ is a non-empty open interval with length $$ m(I \cap J) = \min(b,d) - \max(a, c) > 0 $$ and $$ |a-c|+|b-d| + 2m(I \cap J) \\ = \max(a, c) - \min(a, c) + \max(b, d) - \min(b,d) + 2\bigl(\min(b,d) - \max(a, c)\bigr) \\ = \max(b, d) + \min(b, d) - \max(a, c) - \min(a, c) \\ = (b + d) - (a + c) = (d-a) + (b-c) \\ = |a-d|+|b-c| \, . $$