Suppose $n$ is a positive integer, $x_i \geq 0$ and $\alpha >\beta >0$, is the following inequality true ?
$\left ( \frac{\left ( \sum_{i=1}^{n} x_{i}^{\alpha} \right )^2-\left ( n-1 \right )\sum_{i=1}^{n} x_{i}^{2\alpha}}{n} \right)^{\frac{1}{\alpha}}\leq \left ( \frac{\left ( \sum_{i=1}^{n} x_{i}^{\beta} \right )^2-\left ( n-1 \right )\sum_{i=1}^{n} x_{i}^{2\beta}}{n} \right)^{\frac{1}{\beta}}$
Where we should assume $\left ( \sum_{i=1}^{n} x_{i}^{\alpha} \right )^2-\left ( n-1 \right )\sum_{i=1}^{n} x_{i}^{2\alpha}\geq 0$ and $\left ( \sum_{i=1}^{n} x_{i}^{\beta} \right )^2-\left ( n-1 \right )\sum_{i=1}^{n} x_{i}^{2\beta}\geq 0$
The case for $n=1,2$ is obvious, other cases remain unsolved.
(BTW, try not to use EV theorem to solve this problem)
If you also find it hard to tackle, what if $\alpha=2$ and $\beta=1$ ?
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$\begingroup$ Why "try not to use EV theorem to solve this problem"? $\endgroup$– LSpiceCommented Oct 27 at 23:36
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$\begingroup$ Actually, EV theorem is almost useless in this inequality... $\endgroup$– aftermatherCommented Oct 30 at 14:41
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$\begingroup$ Re, ah, so this is just a helpful suggestion to avoid a dead end rather than a request for a novel proof technique? $\endgroup$– LSpiceCommented Oct 30 at 16:13
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$\begingroup$ As far as I concerned, yes $\endgroup$– aftermatherCommented Oct 30 at 16:23
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