Prove the following inequality, and explain when equality holds: \[ \abs{z+w} \geq \abs{ \abs{z} - \abs{w} } .\]
Solution
Goal: Prove \(\abs{z + w} \ge \abs{\abs{z} - \abs{w}}\) and explain when equality holds.
1. \(\abs{z} \le \abs{z + w} + \abs{w}\).
Triangle inequality applied to \(z = (z + w) + (-w)\): \(\abs{z} \le \abs{z + w} + \abs{-w} = \abs{z+w} + \abs{w}\), hence \(\abs{z} - \abs{w} \le \abs{z + w}\).
2. \(\abs{w} \le \abs{z + w} + \abs{z}\).
Same argument with roles swapped: \(\abs{w} \le \abs{z + w} + \abs{z}\), hence \(\abs{w} - \abs{z} \le \abs{z + w}\).
3. \(\abs{\abs{z} - \abs{w}} \le \abs{z + w}\).
\(\abs{\abs{z} - \abs{w}} = \max(\abs{z} - \abs{w}, \abs{w} - \abs{z}) \le \abs{z + w}\) by 1 and 2.
4. Equality characterization.
2. Conclusion: equality holds iff \(z\) and \(w\) are collinear with opposite directions: \(z/w \in (-\infty, 0]\), i.e. \(\arg z = \arg w + \pi \pmod{2\pi}\) (or one of them is \(0\)).
5. Q.E.D.
3 proves the inequality; 4 characterizes equality.