Mathematics Asked by stranger on December 6, 2020

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.

It is similar to parallelogram identity.

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

Correct answer by Atbey on December 6, 2020

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