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.

  • 1
    $\begingroup$ Of course "defined on $[0,1]$" is not what you want. The preprint says $[0,+\infty)$. $\endgroup$ – Gerald Edgar Apr 18 '18 at 11:24
  • $\begingroup$ @GeraldEdgar: Ah, right! I edited accordingly. $\endgroup$ – Per Alexandersson Apr 18 '18 at 12:49
  • 2
    $\begingroup$ It should follow from Karamata majorization inequality and monotonicity. $\endgroup$ – Fedor Petrov Apr 18 '18 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. $\endgroup$ – Per Alexandersson Apr 18 '18 at 16:04
  • 2
    $\begingroup$ It is the same: $p(p+1)$ terms in both sides after we multiply by $p(p+1)$. $\endgroup$ – Fedor Petrov Apr 18 '18 at 16:41

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.