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Just this little inequality problem. If $c\leq{a}\leq{d}$ and $c\leq{b}\leq{d}$, why is $\mid {a-b} \mid \leq d-c$?

(Used on p. 125 of Rudin's Principles of Mathematical Analysis to prove a property of integration.)

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    $a \le d \land -b \le -c \implies a-b \le d -c$, and $b \le d \land -a \le -c \implies b-a \le d - c$. – 2017-01-07
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    I think the most intuitive way to see this is to draw $a,b,c,d$ on a number line. – 2017-01-07
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    just curious -- how did this question affect my reputation negatively? apparently I have -2 from it but don't see why – 2017-06-13

5 Answers 5

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Without loss of generality, name $a$, $b$ such ad $a \le b$. Now:

$c \le a \le b \le d$

Taken intervals:

$|a-c|+|b-a|+|d-b|=|d-c|$

and, as all previous terms are zero or greater than zero:

$|b-a| \le d-c$

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WLOG assume $a \leq b$. Consider the intervals $[a,b]$ and $[c,d]$.

We have $a \in [c,d]$ and $b \in [c,d]$, thus $[a,b] \subset [c,d]$.

Therefore the diameter of the first interval is smaller than the diameter of the second one.

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    Unless you assumed that $a\leq b$, then its okay – 2017-01-07
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    @juniven fixed ! One may also rename $a$ the minimum of the two and $b$ the maximum. – 2017-01-07
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You've got multiple answers giving analytic arguments, but (as carmichael561 notes in the comments) it's important to remember the conclusion is geometrically obvious:

The length of a subinterval does not exceed the length of the interval

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Without loss of generality, assume $b \le a$, then $c \le b$, thus $-b \le -c$ and $a \le d$, so \begin{align*} |a-b| = a-b \le d-b \le d-c \end{align*}

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We have $$a-b\le d-b\le d-c $$ and $$b-a\le d-a\le d-c $$ hence $$|a-b|=\max\{a-b,b-a\}\le d-c$$