202004241533 Problem 1, Ch. 1 Sec. 1

Prove that if $a$ and $b$ are real numbers then

$||a|-|b||\leqslant |a-b|\leqslant |a|+|b|$ .

Proof.

As $|a|$ is non-negative and $-|a|$ non-positive, one has

Eq. (1):

$-|a|\leqslant a\leqslant |a|$

Eq. (2):

$-|b|\leqslant b\leqslant |b|$

Combining Eq. (1) and Eq. (2),

$-(|a|+|b|)\leqslant a+b\leqslant |a|+|b|$ ,

or, Eq. (3): (the triangle inequality)

$|a+b|\leqslant |a|+|b|$ .

Applying the triangle inequality to $|a-b|$ , one gets

Eq. (4):

$|a-b|=|a+(-b)|\leqslant |a|+|-b|=|a|+|b|$ ,

or, Eq. (4)’:

$|a-b|\leqslant |a|+|b|$ .

Applying the triangle inequality to $|a|=|(a-b)+b|$ , one gets

Eq. (5):

$|a|=|(a-b)+b|\leqslant |a-b|+|b|$ ,

or, Eq. (5)’:

$|a|-|b|\leqslant |a-b|$ .

Applying the triangle inequality to $|b|=|(b-a)+a|$ , one gets

Eq. (6):

$|b|=|(b-a)+a|\leqslant |b-a|+|a|$ ,

Roughwork.

or, Eq. (6)’:

$-|a-b| \leqslant |a|-|b|$

Combining Eq. (5)’ and Eq. (6)’:

$-|a-b| \leqslant |a|-|b| \leqslant |a-b|$ ,

or, Eq. (7):

$||a|-|b||\leqslant |a-b|$ .

Combining Eq. (4)’ and Eq. (7), one obtains readily

$||a|-|b||\leqslant |a-b|\leqslant |a|+|b|$ .

QED