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Let us suppose that $0 \leq p \leq 1$. All variables are assumed to be non-negative.

The function $x \mapsto x^{p+1}$ is strictly convex upwards, so $$ \left( \frac1n \sum_{i=1}^n x_i^{p+1} \right) \geq \left( \frac1n \sum_{i=1}^n x_i \right)^{p+1} $$ with equality iff the $x_i$ are all equal; while $x \mapsto x^p$ is convex downwards, so $$ \left( \frac1n \sum_{i=1}^n x_i^p \right) \leq \left( \frac1n \sum_{i=1}^n x_i \right)^p $$ with equality if the $x_i$ are all equal.

I am not really sure why this is the case. Can anyone explain this more? I try to follow from the definition of convex/concave functions, but it's not working out.

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3 Answers 3

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I try to follow from the definition of convex/concave functions, but it's not working out.

Yes, it is. If the function $u$ is convex, then $$ u\left(\frac1n\sum_{k=1}^nx_k\right)\leqslant\frac1n\sum_{k=1}^nu(x_k). $$ For $u:x\mapsto x^{p+1}$ on $x\geqslant0$ this is your first inequality. For $u:x\mapsto-x^{p}$ on $x\geqslant0$ this is your second inequality.

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  • $\begingroup$ Quite often the definition given for a convex function is the simpler inequality $f(t x + (1-t) y) \le t f(x) + (1-t) f(y)$. $\endgroup$
    – J. J.
    Oct 15, 2013 at 13:19
  • $\begingroup$ @J.J. This might be why the OP is stuck. Yet another example of the "advantage" of putting nothing into a question except its flat statement... $\endgroup$
    – Did
    Oct 15, 2013 at 13:21
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Check out Jensen's inequality. For a convex function $f$ we have $$f\left(\frac{x_1 + \dots + x_n}{n}\right) \le \frac{f(x_1) + \dots + f(x_n)}{n}$$ and the inequality sign is reversed for concave functions.

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A common definition is that $\phi: [a, b] \to \mathbb{R}$ is convex if $\phi(t x + (1-t)y) \leq t\phi(x) + (1-t)\phi(y)$ whenever $0 \leq t \leq 1$ and $x, y \in [a, b]$. Or if you like, $\phi(sx+ty) \leq s\phi(t) + t\phi(y)$ whenever $s+t = 1$ and $s, t \geq 0$.

Now proceed by induction to show that if $t_1 + t_2 + \ldots + t_n = 1$ with $t_i \geq 0$, and if $x_i \in [a, b]$, then $\phi(t_1 x_1 + \ldots + t_n x_n) \leq t_1\phi(x_1) + \ldots + t_n\phi(x_n)$.

Finally, apply this to the situation where $t_i = 1/n$.

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