A complement to Diananda's inequality
A complement to Diananda's inequality
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DOI：
10.7153/mia-2018-21-19
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发表时间：
2018
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影响因子：
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通讯作者：
Peng Gao
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文献类型：
--
作者：
Peng Gao
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 $q=\min q\sb i$..In this paper, we prove analogue inequalities in the reversed direction.
