Proving the reverse triangle inequality of the complex numbers

I'm having trouble understanding this proof for the reverse triangle inequality of the complex numbers.

Suppose for any $z, w \in \mathbb{C}$, we have $|z + w| \leq |z| + |w|$ (the triangle inequality).

Prove $|\,|z| - |w|\,| \leq |z - w|$.

To prove this, we know $|z + w| \leq |z| + |w|$. Then letting $v = z + w$ gives $z = v - w$, and so we get the inequality $|v| \leq |v - w| + |w|$, or equivalently, $|v| - |w| \leq |v - w|$.

Great, so now we have $|v| - |w| \leq |v - w|$. For some reason, this implies that $|w| - |v| \leq |w - v|$, and then this clearly implies $|\,|w| - |v|\,| \leq |w - v|$.

Two questions:

1. We proved that given arbitrary $z, w$ in $\mathbb{C}$, if $v = z + w$, then $|v| - |w| \leq |v - w|$. Why does this imply we can say $|w| - |v| \leq |w - v|$?

2. We only proved the inequality for $v$ of the form $z + w$. How does this magically extend to holding for all complex numbers?

I have a feeling someone will answer 2 with "any arbitrary vector in $\mathbb{C}$ can be expressed as $z + w$ since $z$, $w$ were arbitrary" or something like this. If this is true, please explain further.

• 2. Let $v,w$ arbitrary, then define $z:=v-w$ and apply the triangle inequality. – Berci Oct 24 '14 at 17:29

You know that $|x| \le |x-y|+|y|$ and so $|x|-|y| \le |x-y|$. The same argument with $x,y$ switched gives $|y|-|x| \le |y-x| = |x-y|$. Hence $||x|-|y|| \le |x-y|$.