On a result of Cartwright and Field

• Main Author: Peng Gao
• Format: Article
• Language: English
• Published: SpringerOpen 2018-12-01
• Series: Journal of Inequalities and Applications
• Subjects: Arithmetic-geometric mean; Inequality; Power means
• Online Access: http://link.springer.com/article/10.1186/s13660-018-1948-8

Abstract Let Mn,r=(∑i=1nqixir)1r $M_{n,r}=(\sum_{i=1}^{n}q_{i}x_{i}^{r})^{\frac{1}{r}}$, r≠0 $r\neq 0$, and Mn,0=limr→0Mn,r $M_{n,0}= \lim_{r \rightarrow 0}M_{n,r}$ be the weighted power means of n non-negative numbers xi $x_{i}$, 1≤i≤n $1 \leq i \leq n$, with qi>0 $q_{i} > 0$ satisfying ∑i=1nqi=1 $\sum^{n}_{i=1}q_{i}=1$. For r>s $r>s$, a result of Cartwright and Field shows that when r=1 $r=1$, s=0 $s=0$, r−s2xnσn≤Mn,r−Mn,s≤r−s2x1σn, $$\begin{aligned} \frac{r-s}{2x_{n}}\sigma _{n} \leq M_{n,r}-M_{n,s} \leq \frac{r-s}{2x _{1}} \sigma _{n}, \end{aligned}$$ where x1=min{xi} $x_{1}=\min \{ x_{i} \}$, xn=max{xi} $x_{n}=\max \{ x_{i} \}$, σn=∑i=1nqi(xi−Mn,1)2 $\sigma _{n}= \sum_{i=1}^{n}q_{i}(x_{i}-M_{n,1})^{2}$. In this paper, we determine all the pairs (r,s) $(r,s)$ such that the right-hand side inequality above holds and all the pairs (r,s) $(r,s)$, −1/2≤s≤1 $-1/2 \leq s \leq 1$ such that the left-hand side inequality above holds.