# complex norm inequality

For $$z,w \in \mathbb{C}$$ How to show the identity: $$2(|z|^n+|w|^n) \leq (|z+w|^n + |z-w|^n )?$$ for $$n \geq 2$$. I tried induction, but can't I finish.

• If $z=1$, $w=-1$, $n=1$, don't we get $4 \leq 2$? Nov 11, 2020 at 23:10
• sorry, for $n \geq 2$ Nov 11, 2020 at 23:12
• If you divide by $|z|^n$ then you obtain $$2(1+|\frac{w}{z}|^n)\leq |1+\frac{w}{z}|^n+|1-\frac{w}{z}|^n$$ so you need proof that $$2(1+|z|^n)\leq |1+z|^n+|1-z|^n, \forall z\in \mathbb{C}$$ Nov 11, 2020 at 23:32

\begin{aligned} |z|^n + |w|^n &= ^{n/2} + ^{n/2}\\ &\leq (+)^{n/2} \\&\leq \frac{1}{2^{n/2}}(+)^{n/2}\\&\leq \frac{2^{n/2-1}}{2^{n/2}}(^{n/2}+^{n/2}) = \frac{1}{2}|z+w| +\frac{1}{2} |z-w| \end{aligned}