Here's my working:

$\|\mathbf{u} -\mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + (- 2\, \mathbf{u}\,\bullet\mathbf{v})$

Since, by the Cauchy-Schwarz theorem, $|\mathbf{u}\,\bullet\mathbf{v}| \leq \|\mathbf{u}\| \cdot\|\mathbf{v}\|$,

$-(\mathbf{u} \cdot\mathbf{v})$ is also $\leq \|\mathbf{u}\| \cdot\|\mathbf{v}\|$ thanks to the modulus operator.

So, $\|\mathbf{u} -\mathbf{v}\|^2 \leq \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + 2\,\|\mathbf{u}\| \cdot\|\mathbf{v}\|$

From here onwards, it can be shown that indeed $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\| $ with a little algebra.

So, I seem to have proved that $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\| $, however I am wary of this. Searching in Linear Algebra books, I was not able to find this result anywhere. I found only the "reverse triangle inequality", which is not the same as this. I suspect that I have made a mistake in my proof. Is my proof correct?

  • 3
    $\begingroup$ Your proof is ok. You may have found the triangle inequality $$\lVert x+y\rVert \le \lVert x\rVert +\lVert y\rVert$$ What happens if you set $x=u$ and $y=-v$ ? $\endgroup$ – user228113 May 14 '16 at 17:54
  • $\begingroup$ So are you saying that $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\| $ is the same as the triangle inequality? $\endgroup$ – hb20007 May 14 '16 at 17:55
  • $\begingroup$ Well, yes.${}{}$ $\endgroup$ – user228113 May 14 '16 at 17:56
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    $\begingroup$ Since $\lVert -v\rVert = \lVert v\rVert$, yes. $\endgroup$ – Clement C. May 14 '16 at 17:56
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    $\begingroup$ an interesting question is : if $\|cu\| = |c| \ \|u\|$, $\ \|u\| \ge 0$, $\ \|u\|=0 \implies u=0$ and $\|u+v\|^2 \le \|u\|^2+\|v\|^2 + 2 \|u\| \|v\|$, do we get $\|u+v\| \le \|u\| + \|v\|$ i.e. that $\|.\|$ is a norm ? en.wikipedia.org/wiki/Norm_(mathematics)#Definition $\endgroup$ – reuns May 14 '16 at 19:12

The proof is correct.

$\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\|$ is the same as the triangle inequality: $\|\mathbf{x} +\mathbf{y}\|\leq \|\mathbf{x}\| + \|\mathbf{y}\|$.

Since $\|-\mathbf{v}\|= \|\mathbf{v}\|$.

Full credit goes to G. Sassatelli and Clement C. for providing the full answer in comments.

Answer derived from their comments based on SE policy here.


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