English

A complement to Diananda's inequality

Classical Analysis and ODEs 2016-11-29 v1

Abstract

Let Mn,r=(i=1nqixir)1r,r0M_{n,r}=(\sum_{i=1}^{n}q_ix_i^r)^{\frac {1}{r}}, r \neq 0 and Mn,0=limr0Mn,rM_{n,0}=\lim_{r \rightarrow 0}M_{n,r} be the weighted power means of nn non-negative numbers xix_i with qi>0q_i > 0 satisfying i=1nqi=1\sum^n_{i=1}q_i=1. In particular, An=Mn,1,Gn=Mn,0A_n=M_{n,1}, G_n=M_{n,0} are the arithmetic and geometric means of these numbers, respectively. A result of Diananda shows that \begin{align*} M_{n,1/2}-qA_n-(1-q)G_n & \geq 0,\\ M_{n,1/2}-(1-q)A_n-qG_n & \leq 0,\end{align*} where q=minqiq=\min q_i. In this paper, we prove analogue inequalities in the reversed direction.

Keywords

Cite

@article{arxiv.1611.09126,
  title  = {A complement to Diananda's inequality},
  author = {Peng Gao},
  journal= {arXiv preprint arXiv:1611.09126},
  year   = {2016}
}

Comments

10 pages

R2 v1 2026-06-22T17:06:22.316Z