Prove the following inequality, and explain when equality holds: \[ \abs{z-w} \geq \abs{ \abs{z} - \abs{w} } .\]
Solution
\[ \abs{z-w}^2 &= (z-w)(\bar z - \bar w) \\ &= \abs{z}^2 + \abs{w}^2 - z\bar{w} - \bar{z} w \\ &= \abs{z}^2 + \abs{w}^2 - 2\Re(\bar w z) \\ &\geq \abs{z}^2 + \abs{w}^2 - 2\abs{\bar w }\abs{z} \\ &\geq \qty{ \abs{z} - \abs{w} }^2 ,\] and taking square roots introduces an absolute value on the final term. Here we’ve used the basic estimate \[ \Re(z) \leq \abs{z} \implies -\Re(z) \geq -\abs{z} .\]