# 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

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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*}

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pharmine, Thank you. – VVV Nov 13 '11 at 22:00
You're welcome. – pharmine Nov 15 '11 at 6:23

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" ?

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