On Kedlaya-type inequalities for weighted means

Journal of Inequalities and Applications

Paweł Pasteczka¹  Zsolt Páles² 

[1] Institute of Mathematics, Pedagogical University of Cracow;Institute of Mathematics, University of Debrecen;

关键词: Discrete mean;    Weighted mean;    Power mean;    Quasi-arithmetic mean;    Gini mean;    Deviation mean;   

DOI  :  10.1186/s13660-018-1685-z

来源: DOAJ

【 摘 要 】

Abstract In 2016 we proved that for every symmetric, repetition invariant and Jensen concave mean M $\mathscr{M}$ the Kedlaya-type inequality A(x1,M(x1,x2),…,M(x1,…,xn))≤M(x1,A(x1,x2),…,A(x1,…,xn)) $$ \mathscr{A} \bigl(x_{1},\mathscr{M}(x_{1},x_{2}), \ldots,\mathscr{M}(x _{1},\ldots,x_{n}) \bigr) \le \mathscr{M} \bigl( x_{1}, \mathscr{A}(x _{1},x_{2}), \ldots,\mathscr{A}(x_{1},\ldots,x_{n}) \bigr) $$ holds for an arbitrary (xn) $(x_{n})$ ( A $\mathscr{A}$ stands for the arithmetic mean). We are going to prove the weighted counterpart of this inequality. More precisely, if (xn) $(x_{n})$ is a vector with corresponding (non-normalized) weights (λn) $(\lambda_{n})$ and Mi=1n(xi,λi) $\mathscr{M}_{i=1}^{n}(x _{i},\lambda_{i})$ denotes the weighted mean then, under analogous conditions on M $\mathscr{M}$, the inequality holds for every (xn) $(x_{n})$ and (λn) $(\lambda_{n})$ such that the sequence (λkλ1+⋯+λk) $(\frac{\lambda_{k}}{\lambda_{1}+\cdots+\lambda_{k}})$ is decreasing.

【 授权许可】

Unknown