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Popoviciu's inequality states that for convex $f$ and numbers $x_1,x_2,x_3$, we have $f(x_1)+f(x_2) + f(x_3) + 3\cdot f(\frac{x_1+x_2+x_3}3) \geq 2\cdot f(\frac{x_1+x_2}2)+2\cdot f(\frac{x_1+x_3}2)+2\cdot f(\frac{x_2+x_3}2)$.

It can be proven just by using Karamata's inequality and arguing that the sequence of $6$ numbers on the LHS majorizes the sequence on the RHS. Some modest generalizations were proposed, where we would have $n$ numbers, LHS would have still a sum of the function for each single argument + the mean and the RHS would have average of all possible $k$ of $n$ tuples.

Can we go full measure here with all possible combinations of numbers? That is, $\sum f(x_i) - 2 \sum_{i_1 < i_2 } f(\frac{x_{i_1}+x_{i_2}}{2})+ 3 \sum_{i_1 < i_2 < i_3} f(\frac{x_{i_1}+x_{i_2}+x_{i_3}}{3}) + ... + (-1)^{n+1} nf(\frac{x_{i_1}+...+x_{i_n}}{n}) \geq 0$

I would like to tackle this with Karamata's inequality and argue that the sequence for positive numbers majorizes the sequence for negative numbers. However already for $n=3$ the only way to show the majorization is with case work. So I would like to know if there are some smarter ways of doing so. Or maybe you know some obvious reasons for which the whole inequality just cannot be true.

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  • $\begingroup$ It might be worth numerical calculations to verify that it seems to be true... $\endgroup$
    – Joseph O'Rourke
    Commented Jun 28, 2015 at 19:16
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    $\begingroup$ I fear this is not true. See my post at artofproblemsolving.com/community/c6h244828p1348667 $\endgroup$
    – darij grinberg
    Commented Jun 28, 2015 at 19:24
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    $\begingroup$ There are some analogues of Popoviciu's inequality which do have a more interesting structure than just values and their average on the LHS. See artofproblemsolving.com/community/c6h98832p557719 $\endgroup$
    – darij grinberg
    Commented Jun 28, 2015 at 19:26
  • $\begingroup$ Dear Dr. @darijgrinberg I give a conjecture generalization of Popociviu's inequality as following, and I am looking for a proof. Let $f(x)$ is a real continuous function that is strictly convex on $[m, M]$, let $m \le x_1 \le x_2 \le x_3 \le...\le x_n \le M$ then show that: $nf\left(\frac{x_1+\cdots+x_n}{n}\right)+f(x_1)+\cdots+f(x_n) \ge 2f(\frac{x_1+x_2}{2})+....+2f(\frac{x_{n-1}+x_n}{2})+2f(\frac{x_{n}+x_1}{2}) $ Equality holds if only if $x_1=x_2=\cdots=x_n$ $\endgroup$
    – Oai Thanh Đào
    Commented Aug 25, 2016 at 14:02

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Let me elaborate on Darij's comment: Convex functions on a closed interval can be uniformly approximated by piecewise linear convex functions, and it is a theorem of Popoviciu that piecewise linear convex functions are of the form $$f(x)=ax+b+\sum c_i\left|x-r_i\right|,$$ where the $c_i$ are non-negative.

With this in mind, inequalities like Popoviciu's always can be checked by looking at the corresponding "abslute value" version. In this case it follows from the (more familiar to some people) Hlwaka inequality: $$|x|+|y|+|z|+|x+y+z|\geq |x+y|+|y+z|+|x+z|.$$ Denis Serre asked a question equivalent to yours a few months ago, about an n-variable Hlawka inequality, but the same counterexample was pointed out.

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  • $\begingroup$ Dear Dr. @GjergjiZaimi I give a conjecture generalization of Popociviu's inequality as following, and I am looking for a proof. Let $f(x)$ is a real continuous function that is strictly convex on $[m, M]$, let $m \le x_1 \le x_2 \le x_3 \le...\le x_n \le M$ then show that: $nf\left(\frac{x_1+\cdots+x_n}{n}\right)+f(x_1)+\cdots+f(x_n) \ge 2f(\frac{x_1+x_2}{2})+....+2f(\frac{x_{n-1}+x_n}{2})+2f(\frac{x_{n}+x_1}{2}) $ Equality holds if only if $x_1=x_2=\cdots=x_n$ $\endgroup$
    – Oai Thanh Đào
    Commented Aug 25, 2016 at 13:57

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