Prove for any vector norm $\| \cdot \|$ that $\left| \|x\| -\|y\| \right| \leq \|x - y\|$

Question

To prove that for any vector norm $\| \cdot \|$ that $\left| \|x\| - \|y\| \right| \leq \|x - y\|$ I've been trying to follow the derivation found here http://fourier.eng.hmc.edu/e161/lectures/algebra/node11.html but I can only understand up to the step where we get $\|x - z\| \leq \|x\| + \|z\|$ but I get stuck on how subtracting $\|y\|$ from both sides then defining $z = z + y$ gets us $\|z\| - \|y\| \leq \|z - y\|$. Any help is appreciated.

Edit: The proof found can be done using the following 3 definitions: (1) $\|x\| \geq 0$, (2) $\|\alpha x \| = |\alpha| \|x\|$, and (3) the triangle inequality $\|x + y\| \leq \|x\| + \|y\|$.

Answer 1

You have the wrong substitution. Try $x=z-y$.

Answer 2

Note that $\|x-z\| \le \|x\| + \|z\|$ holds for all values of $x$ and $z$, so if we let $z = x+y$ and subtract $\|y\|$ from both sides, we get $$ \begin{split} \|x-z\| &\le \|x\| + \|z\|\\ \|x-(x+y)\| - \|y\| &\le \|x\| + \|(x+y)\| - \|y\|\\ \|-y\| - \|y\| &\le \|x\| + \|(x+y)\| - \|y\|\\ 0 &\le \|x\| + \|(x+y)\| - \|y\|\\ \end{split} $$ and now the result follows.

Answer 3

You have $\|a+b\| \le \|a\| + \|b\|.$

So $\|a+b\| - \|a\| \le \|b\|.$

Letting $a=x$ and $b=y-x,$ we get $$ \| x + (y-x) \| -\|x\| \le \| y-x\| $$ $$ \|y\|-\|x\| \le \|y-x\| $$ Similarly we get $\|x\|-\|y\| \le \|x-y\| = \|y-x\|.$