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.
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6$\begingroup$ If $z=1$, $w=-1$, $n=1$, don't we get $4 \leq 2$? $\endgroup$– AtbeyNov 11, 2020 at 23:10
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1$\begingroup$ sorry, for $n \geq 2$ $\endgroup$– strangerNov 11, 2020 at 23:12
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1$\begingroup$ 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 $$ 2(1+|z|^n)\leq |1+z|^n+|1-z|^n, \forall z\in \mathbb{C} $$ $\endgroup$– ZhoooNov 11, 2020 at 23:32
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1 Answer
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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}
