Editing Triangle inequality (section)

==Reverse triangle inequality==
The '''reverse triangle inequality''' is an equivalent alternative formulation of the triangle inequality that gives lower bounds instead of upper bounds. For plane geometry, the statement is:<ref name=inequality>
{{cite book |title=The popular educator; fourth volume |chapter-url=https://books.google.com/books?id=lTACAAAAQAAJ&pg=PA196 |page=196 |chapter=Exercise I. to proposition XIX |year=1854 |publisher=John Cassell |location=Ludgate Hill, London |author=Anonymous}}
</ref>

:''Any side of a triangle is greater than or equal to the difference between the other two sides''.

In the case of a normed vector space, the statement is:
: <math>\big|\|u\|-\|v\|\big| \leq \|u-v\|,</math>
or for metric spaces, <math>|d(A, C) - d(B, C)| \leq d(A, B)</math>.
This implies that the norm <math>\|\cdot\|</math> as well as the distance-from-<math>z</math> function <math>d(z ,\cdot)</math> are [[Lipschitz continuity|Lipschitz continuous]] with Lipschitz constant {{math|1}}, and therefore are in particular [[uniform continuity|uniformly continuous]].

The proof of the reverse triangle inequality from the usual one uses <math> \|v-u\| = \|{-}1(u-v)\| = |{-}1|\cdot\|u-v\| = \|u-v\| </math> to find:
: <math> \|u\| = \|(u-v) + v\| \leq \|u-v\| + \|v\| \Rightarrow \|u\| - \|v\| \leq \|u-v\|, </math>
: <math> \|v\| = \|(v-u) + u\| \leq \|v-u\| + \|u\| \Rightarrow \|u\| - \|v\| \geq -\|u-v\|, </math>

Combining these two statements gives:
: <math> -\|u-v\| \leq \|u\|-\|v\| \leq \|u-v\| \Rightarrow \big|\|u\|-\|v\|\big| \leq \|u-v\|.</math>
In the converse, the proof of the triangle inequality from the reverse triangle inequality works in two cases:

If <math>\|u +v\| - \|u\| \geq 0,</math> then by the reverse triangle inequality, <math>\|u +v\| - \|u\| = {\big|}\|u + v\|-\|u\|{\big|} \leq \|(u + v) - u\| = \|v\| \Rightarrow \|u + v\| \leq \|u\| + \|v\|</math>,

and if <math>\|u +v\| - \|u\| < 0,</math> then trivially <math>\|u\| +\|v\| \geq \|u\| > \|u + v\|</math> by the nonnegativity of the norm. 

Thus, in both cases, we find that <math>\|u\| + \|v\| \geq \|u + v\|</math>.

For metric spaces, the proof of the reverse triangle inequality is found similarly by:

<math>d(A, B) + d(B, C) \geq d(A, C) \Rightarrow d(A, B) \geq d(A, C) - d(B, C)</math>

<math>d(C, A) + d(A, B) \geq d(C, B) \Rightarrow d(A, B) \geq d(B, C) - d(A, C)</math>

Putting these equations together we find:

<math>d(A, B) \geq |d(A, C) - d(B, C)|</math>

And in the converse, beginning from the reverse triangle inequality, we can again use two cases:

If <math>d(A, C) - d(B, C) \geq 0</math>, then <math>d(A, B) \geq |d(A, C) - d(B, C)| = d(A, C) - d(B, C) \Rightarrow d(A, B) + d(B, C) \geq d(A, C)</math>,

and if <math>d(A, C) - d(B, C) < 0,</math> then <math>d(A, B) + d(B, C) \geq d(B, C) > d(A, C)</math> again by the nonnegativity of the metric. 

Thus, in both cases, we find that <math>d(A, B) + d(B, C) \geq d(A, C)</math>.