Q:

Showing that the magnitude of the difference of two vectors is larger than the difference of it's vector magnitudes

Long title. I have to prove (the problem itself suggests using Pythagorean theorem) the following inequality:

$$\|u\|-\|v\| \le \|u-v\| $$

Vector magnitudes... How do you prove this in an eloquent fashion?
Edited out dot product, translating algebra is really hit or miss sometimes. 

A:

Let's prove something else instead.  Let's prove ${\big |}\|u\|-\|v\|{\big |} \le \|u-v\|$.
Square the LHS to get: $(\|u\|-\|v\|)^2 = \|u\|^2-2\|u\|\|v\|+\|v\|^2$.
Then we know from the Cauchy inequality that $(\|u\|-\|v\|)^2 = \|u\|^2-2\|u\|\|v\|+\|v\|^2 \le \|u\|^2-2(u\cdot v)+\|v\|^2 =\|u-v\|^2$
Take the square root of both sides to get ${\big |}\|u\|-\|v\|{\big |} \le \|u-v\|$.
Because $a\le |a|$ for all $a\in \Bbb R$, we have $\|u\|-\|v\| \le {\big |}\|u\|-\|v\|{\big |} \le \|u-v\|$.  $\ \ \ \square$
Thus by proving this stricter inequality, we get yours back for free.

