So, we managed to prove the following inequality: Let $f$ be a decreasing convex function on $\mathbb{R}_{\geq 0}$, let $p$ be a positive integer and $r\geq 0$. Then \begin{align} \frac{1}{p} \sum_{\ell=1}^{p} f\left(\frac{\ell+r}{p} \right) \leq \frac{1}{p+1} \sum_{\ell=1}^{p+1} f\left( \frac{\ell+r}{p+1} \right). \end{align}

Does this inequality have a name? Is it known?

It is related to this question over at math.stackexchange.

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    $\begingroup$ Of course "defined on $[0,1]$" is not what you want. The preprint says $[0,+\infty)$. $\endgroup$ Apr 18, 2018 at 11:24
  • $\begingroup$ @GeraldEdgar: Ah, right! I edited accordingly. $\endgroup$ Apr 18, 2018 at 12:49
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    $\begingroup$ It should follow from Karamata majorization inequality and monotonicity. $\endgroup$ Apr 18, 2018 at 15:50
  • $\begingroup$ @FedorPetrov Yes, I thought so too, but note that the number of terms in the LHS is not the same as RHS. We did not manage to derive the above as a consequence. $\endgroup$ Apr 18, 2018 at 16:04
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    $\begingroup$ It is the same: $p(p+1)$ terms in both sides after we multiply by $p(p+1)$. $\endgroup$ Apr 18, 2018 at 16:41

1 Answer 1


So, the proof is available now in the appendix of this preprint, and we found a few references to the case $r=0$.


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