# Is $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\|$?

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.

Is my proof correct?

• 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 $y=-v$ ?
– user228113
Commented May 14, 2016 at 17:54
• So are you saying that $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\|$ is the same as the triangle inequality? Commented May 14, 2016 at 17:55
• Well, yes.${}{}$
– user228113
Commented May 14, 2016 at 17:56
• Since $\lVert -v\rVert = \lVert v\rVert$, yes. Commented May 14, 2016 at 17:56
• 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 Commented May 14, 2016 at 19:12

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