prove $\sum^n_{i=1} x_i^{-p} \ge n (\bar x)^{-p} $

Question

Can we prove that the sum of negative powers is larger than n times of the negative power of the mean? i.e. with $x_i > 0$, $p \geq 0$,

$$\sum^n_{i=1} x_i^{-p} \ge n (\bar x)^{-p} $$ in which $\bar x = \frac{1}{n}\sum^n_{i=1} x_i $ is the mean.

Numerical tests show this inequality holds. Is there a theoretical proof of this inequality? Thanks!

Answer 1

In terms of expectations, your claim is $E(X^{-p})\ge(EX)^{-p}$. By Jensen's theorem, this follows if $p\ge0$ or $p\le-1$, since $(x^{-p})^{\prime\prime}=p(p+1)x^{-p-2}$.