  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.
