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Prove the pth power mean is less than the (2p)th power mean

by RoRi

We recall the definition of the pth power mean M_p.

For x_1, \ldots, x_n \in \mathbb{R}_{>0}, and p \in \mathbb{Z} with p \neq 0, we define the pth power-mean M_p as:

    \[ M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}. \]

Now, for p > 0, prove M_p < M_{2p} for x_1, \ldots, x_n not all equal.

Proof. From the Cauchy-Schwarz inequality we know that for real numbers a_1, \ldots, a_n and b_1, \ldots, b_n, we have

    \[ \left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right) \]

with equality if and only if there is some y \in \mathbb{R} such that a_k y + b_k = 0 for all k. Letting a_k = x_k^p and b_k = 1 we have

    \[ \left( \sum_{k=1}^n x_k^p \cdot 1 \right)^2 < \left( \sum_{k=1}^n x_k^{2p} \right) \left( \sum_{k=1}^n 1 \right). \]

This inequality is strict since if equality held there would exist some y \in \mathbb{R} such that (x_k^p)y + 1 = 0 for all k, but this would imply x_k = \left(-\frac{1}{p} \right)^{1/p} for all k, contradicting our assumption that the x_k are not all equal. Since \sum_{k=1}^n 1 = n (see here), this implies

    \begin{align*} &&\sum_{k=1}^n x_k^p &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2} n^{1/2} \\ \implies && \left( \sum_{k=1}^n x_k^p \right)^{1/p} &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2p} \cdot n^{1/2p} &(\text{raising to } 1/p)\\ \implies && \left (\frac{1}{n} \right)^{1/p} \cdot \left( \sum_{k=1}^n x_k^p \right)^{1/p} &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2p} \cdot \left( \frac{1}{n} \right)^{1/2p} & (\text{multiplying by } (1/n)^{1/p})\\ \implies && \left( \frac{\sum_{k=1}^n x_k^p}{n} \right)^{1/p} &< \left( \frac{\sum_{k=1}^n x_k^{2p}}{n} \right)^{1/2p} & (\text{combining terms})\\ \implies && M_p &< M_{2p}. \qquad \blacksquare \end{align*}

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