Proving a Vector Norm Inequality Holds

Suppose we have a vector norm defined as $v(x)$ for $x \in\Bbb{ R}^n$, show that for vectors $x,y \in \Bbb{R}^n$ the following inequality holds:

$|v(x) - v(y)| \leq v(x - y)$.

I know that if $v(x)$ is a vector norm, then one of the conditions it necessarily satisfies is:

$||x + y|| \leq ||x|| + ||y||$ for all $x,y \in R^n$.

I believe the proof of the above inequality will most likely make use of this particular property but want to be careful when specifically using it and am having trouble establishing it. So far the attempt at showing the inequality I have is:

$|v(x) - v(y)| = ||x| - |y|| \leq |x| - |y|$

I am confused whether I have properly used the inequality property of vector norms and have trouble showing the original inequality, specifically how to get to the RHS of inequality. Any insight or help would be much appreciated, thanks.

Answer

You can use the triangle inequality as follows:
$$
\begin{align}
v(x) = v(y + (x-y)) \leq v(y) + v(x-y) \implies v(x)-v(y)\leq v(x-y)\tag1\\
v(y) = v(x + (y-x)) \leq v(x) + v(y-x) \implies v(y)-v(x)\leq v(y-x)\tag2

\end{align}
$$
thus
$$
\begin{align}
v(x)-v(y)\leq v(x-y)\tag1\\
-(v(x)-v(y))\leq v(x-y)\tag2
\end{align}
$$

using (1) and (2) $|v(x)-v(y)|\leq v(x-y)$.