Show that $(\frac{1}{n} \sum_{i=1}^n |x_i|^p)^{1/p} \ge \frac{1}{n} \sum_{i=1}^n |x_i|$

Question

Question: Show that $(\frac{1}{n} \sum_{i=1}^n |x_i|^p)^{1/p} \ge \frac{1}{n} \sum_{i=1}^n |x_i|$ in which $p>1$, $x_i \in \mathbb{R}$, $i=1,2,\dots,n$

Could you give me some hint to solve this problem. I see that it like the Minkovski's inequality but it not true. Thank all!

Google Power Means Inequality... – Macavity Jul 10 '19 at 13:00 — Macavity, Jul 10 '19 at 13:00

score 0 · Answer 1 · answered Jul 10 '19 at 11:21

0

It can be easily showed by Weighted Power Mean. $\;$ You can see Weighted Power Mean by here on page $2$ https://artofproblemsolving.com/articles/files/MildorfInequalities.pdf

answered Jul 10 '19 at 11:21

Taha Direk

1,220

Show that $(\frac{1}{n} \sum_{i=1}^n |x_i|^p)^{1/p} \ge \frac{1}{n} \sum_{i=1}^n |x_i|$

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