On Kedlaya type inequalities for weighted means

Abstract

In 2016 we proved that for every symmetric, repetition invariant and Jensen concave mean $\mathscr{M}$ the Kedlaya-type inequality $$\mathscr{A}\big(x_1,\mathscr{M}(x_1,x_2),\ldots,\mathscr{M}(x_1,\ldots,x_n)\big)\le \mathscr{M} \big(x_1, \mathscr{A}(x_1,x_2),\ldots,\mathscr{A}(x_1,\ldots,x_n)\big)$$ holds for an arbitrary $(x_n)$ ($\mathscr{A}$ stands for the arithmetic mean). We are going to prove the weighted counterpart of this inequality. More precisely, if $(x_n)$ is a vector with corresponding (non-normalized) weights $(\lambda_n)$ and $\mathscr{M}_{i=1}^n(x_i,\lambda_i)$ denotes the weighted mean then, under analogous conditions on $\mathscr{M}$, the inequality $$\mathscr{A}_{i=1}^n \big(\mathscr{M}_{j=1}^i (x_j,\lambda_j),\:\lambda_i\big) \le \mathscr{M}_{i=1}^n \big(\mathscr{A}_{j=1}^i (x_j,\lambda_j),\:\lambda_i\big)$$ holds for every $(x_n)$ and $(\lambda_n)$ such that the sequence $(\frac{\lambda_k}{\lambda_1+\cdots+\lambda_k})$ is decreasing.Comment: J. Inequal. Appl. (2018