# Norm inequality on Complex Numbers.

For $z,w \in \mathbb{C}$, it is true that $2 | z w| \leq |z|^2 + |w|^2$. How does this imply the identity: $$|z+w|^2 \leq 2(|z|^2 + |w|^2 )?$$

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Edited title to make it "inequality" rather than "identity." An identity is an equality - think the related word, "identical." –  Thomas Andrews Apr 26 '12 at 13:21

$$|z+w|^2=|z^2+w^2+2zw|\leq|z|^2+|w|^2+2|zw|\leq 2(|z|^2+|w|^2)$$

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Or from just substituting the inequality Peter gave. –  Alex Becker Apr 26 '12 at 13:30

\begin{eqnarray*} | z + w|^{2} & = & (z+w)(\overline{z+w})\\ & = & (z+w)(\overline{z} + \overline{w})\\ & = & z \cdot \overline{z} + z \overline{w} + w \overline{z} + w \overline{w} \\ & \le & |z|^{2} + |w|^{2} +|z \overline{w}| + |w \overline{z}|\\ & = & |z|^{2} + |w|^{2} + 2|z w|\\ & \le & 2(|z|^{2} + |w|^{2})\\ \end{eqnarray*}

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