Elegant way to prove this inequality 
$1-|w| \le |1+w| \le 1+|w| \ \forall w \in \mathbb{C}$  

The book gives a geometric  but not an algebraic proof. Does anybody see a way?  Please do tell 
 A: Suppose you can assume the triangle inequality
$$|a+b|\le |a|+|b|\quad \text{for}\ a,\,b\in\mathbb{C}.$$
By letting $a=-w$ and $b=1+w$ we get
$|1|\le |-w|+|1+w|$, hence $1-|w|\le |1+w|$;
By letting $a=1$ and $b=w$ we get
$|1+w|\le |1|+|w|$, hence $|1+w|\le 1+|w|$.
Edit. "Algebraic proof" of the triangle inequality.
We show that $(|a|+|b|)^2-(|a+b|)^2\ge 0$.
$$\begin{align*}
(|a|+|b|)^2-(|a+b|)^2 &=|a|^2+2|a||b|+|b|^2-(a+b)(\overline{a+b})\\
&=a\overline{a}+2|a||b|+b\overline{b}-a\overline{a}-b\overline{b}
-a\overline{b}-\overline{a}b\\
&=2|a||\overline{b}|-(a\overline{b}+\overline{a\overline{b}})\\
&=2|a\overline{b}|-2\text{Re}\,(a\overline{b})\ge 0.
\end{align*}$$
A: You don't need geometry, this is just basic properties of norms : The triangle inequality $|x+y|\leq |x|+|y|$ yields your upper bound by taking $x=1$, and you obtain your lower bound by using $||x|-|y||\leq |x-y|$ and the trivial fact that $|x|-|y|\leq||x|-|y||$ specifying $x$ to be $1$. But maybe I didn't understand what you mean by "algebraic proof" ?
