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arXiv · 1611.09126

A complement to Diananda's inequality

Peng Gao
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Abstract

Let $M_{n,r}=(\sum_{i=1}^{n}q_ix_i^r)^{\frac {1}{r}}, r \neq 0$ and $M_{n,0}=\lim_{r \rightarrow 0}M_{n,r}$ be the weighted power means of $n$ non-negative numbers $x_i$ with $q_i > 0$ satisfying $\sum^n_{i=1}q_i=1$. In particular, $A_n=M_{n,1}, G_n=M_{n,0}$ 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 $q=\min q_i$. In this paper, we prove analogue inequalities in the reversed direction.

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Peng Gao. 2016-11-28. A complement to Diananda's inequality. https://arxiv.org/abs/1611.09126

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