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If $x_1,\dots,x_n$ are positive numbers, then we can define varies means by $$M_{\mu}(x_1,\dots,x_n) = \left(\frac{x_1^\mu+\dots+x_n^\mu}{n}\right)^{1/\mu}. $$ Then, it is known that $$M_{\mu}(x_1,\dots,x_n) \leq M_{\eta}(x_1,\dots,x_n)$$ for all $\mu\leq \eta$.

My question is does the same hold if we allow $x_1,\dots,x_n$ to be complex?

Obviously this needs to be modified slightly. We'll say $M_{\mu}(x_1,\dots,x_n)$ is well defined if $x_1^\mu+\dots+x_n^\mu\not=0$. Then does something like $$|M_{\mu}(x_1,\dots,x_n)| \leq |M_{\nu}(x_1,\dots,x_n)|$$ as long as $M_{\mu}(x_1,\dots,x_n)$ and $M_{\eta}(x_1,\dots,x_n)$ are well defined?

If not, is there some other construction of means that would give similar results?

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Counterexample (note that negative real numbers are a special case of complex numbers):

$n=2$, $x_1=1$, $x_2=-1$, $\mu=2$, $\nu=3$.

Clearly $\mu=2<\nu=3$. But $$\left|M_2(1,-1)\right| = \left|\left(\frac{1^2+(-1)^2}{2}\right)^{1/2}\right| = 1 > 0 = \left|\left(\frac{1^3+(-1)^3}{2}\right)^{1/3}\right| = \left|M_3(1,-1)\right|$$

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