Let $k,m,n\in\mathbb N$ such that $n>k$. For a partition $\alpha=(\alpha_1,\dots,\alpha_k)\vdash m$ with $\alpha_1\ge\dots\ge \alpha_k>0$ and nonnegative $ x_1,\dots,x_n$ define $x^\alpha :=\prod_{i=1}^k x_i^{\alpha_i} $ and ${\tilde x^{\alpha}} :=\prod_{i=1}^k x_{i+1}^{\alpha_i} $. All indices are cyclical $\mod n$.

Prove or disprove: $${\prod_{cyc}(x_1^m+ \tilde x^{\alpha} )\ge\prod_{cyc}(x_1^m+ x^{\alpha})}.$$

For exemple, $\alpha=(1,1)$ yields the inequality $$\prod_{cyc}(x_{i}^{2}+x_{i+1}x_{i+2})\ge \prod_{cyc}(x_{i}^{2}+x_{i}x_{i+1}),$$ or in more sloppy but intuitive notation, if the variables are denoted $a,b,c,...,z$, $$\prod_{cyc}(a^2+bc)\ge \prod_{cyc}(a^2+ab ).$$

A two-line proof for this:

By Cauchy-Schwarz, $ (a^2 + bc)(c+b) \ge (a\sqrt {c} + \sqrt {bc}\sqrt {b})^2 = c(a + b)^2$.
Now multiply all cyclic permutations of that and note that $\prod\limits_{cyc}(c+b)=\prod\limits_{cyc}(a + b)$.

More generally, for a partition $(p,q)$ with $p\ge q$, putting $$x=\frac{q}{p+q},y=\frac{p-q}{2(p+q)}, z=\frac12,$$ we have $x+y+z=1$ and can apply the discrete form of Hölder's inequality to obtain $$(a^{p+q}+b^pc^q)^x(b^q+c^q)^y(c^q+b^q)^z\ge a^{(p+q)x}b^{qy}c^{qz}+b^ {px+qz} c^{qx+qy}=(a^q+b^q)b^{qy} c^{qz} ,$$ which after multiplying all its cyclic permutations and dividing by $\prod\limits_{cyc}(a^q+b^q)^{y+z}$, leaves us with $$\prod\limits_{cyc} (a^{p+q}+b^pc^q)^x \ge \prod\limits_{cyc}(a^q+b^q)^{x}\prod\limits_{cyc}a^{q(y+z)}=\prod\limits_{cyc}(a^q+b^q)^{x}\prod\limits_{cyc}a^{px} ,$$ qed.

This proof settles $k=2$. (The inequality is wrong for $p<q$). For $k\ge3$ and some special choices of $\alpha$, there are elegant proofs using Cauchy-Schwarz or Hölder. But not for the general case.

A bit of motivation: Looking at some famous inequalities of discrete variables like the power means inequality or Maclaurin's inequality, but also Cauchy-Schwarz, a recurrent principle is that for homogenous inequalities, the more "separated" the variables are, the bigger the expression, and the more "mixed" they are (in the sense of multiplied with each other), the smaller the expression. This coincides with the patterns in question here, giving some "empirical support" in favour of the conjecture.

Even though the condition of non-increasing parts of $\alpha\vdash m$ does not seem always necessary for $k\ge 3$, I think the above conjecture is very tough. (Of course I would be happy to learn the contrary.) I don't even know of a proof for the simplest case of $k=3$, viz. $\alpha=(1,1,1)$, yielding $$\prod_{cyc}(a^3+bcd)\ge \prod_{cyc}(a^3+abc )$$ for $n\ge4$ variables. It might be possible to use some kind of majorization, but I don't see how.
Based on strong numerical evidence, I also have a different "iff" conjecture for $k=3$:

Conjecture: for a partition $\alpha=(p,q,r)$, the inequality $${\prod_{cyc}(x_1^m+ \tilde x^{\alpha} )\ge\prod_{cyc}(x_1^m+ x^{\alpha})}$$ holds for all nonnegative $ x_1,\dots,x_n$ iff both $p\ge r$ and $p+r\ge q$ hold.
It seems tempting to conjecture for each $k\ge4$ the existence of an "iff" statement with a similar set of conditions.

As a side note: In an AoPS thread which deals with a somewhat dual question in 3 variables (the question: for which pairs of partitions $(u,v,w),(p,q,r)\vdash m$ does the inequality $\sum\limits_{cyc} a^ub^vc^w\geq \sum\limits_{cyc} a^{p}b^{q}c^{r}$ hold $ \forall a,b,c\ge0$ ?), it turned out that this holds iff it can be reduced to the weighted AM-GM inequality in a similar way as the reduction to Hölder above.


This is an idea/a partial answer.

Let $x_1,\dots,x_n$ be positive nubmers such that $\prod x_i=1$. Assume that $y_i=\prod_{k=0}^{n-1} x_{i+k}^{\lambda_k}$ for some exponents $\lambda_k$, indices are modulo $n$. Note that we may replace $\lambda_k$ to $\lambda_k+c$ for any constant $c$. Now assume that $\sum |\lambda_i+c|\leq 1$ for some $c$. Then we may choose $c$ such that $\sum |\lambda_i+c|=1$ (by continuity). Now apply Jensen inequality for the convex function $\log(1+e^x)$: Denote $\beta_k=|\lambda_k+c|$, $z_{i,k}=x_{i+k}^{{\rm sign}\,(\lambda_k+c)}$, $$ \prod (1+y_i)=\prod_i \left(1+\prod_{k=0}^{n-1} z_{i,k}^{\beta_k}\right)\leqslant \prod_i \prod_k (1+z_{i,k})^{\beta_k}=\prod (1+x_i), $$
since $\prod (1+1/x_i)=\prod_j (1+x_i)=\prod_i (1+z_{i,k})$ for any fixed $k$.

So, if $\min_c \sum_i |\lambda_i+c|\leq 1$, we get $\prod (1+y_i)\leqslant \prod (1+x_i)$. Apply it to, say $\prod (a^3+bcd)\geqslant \prod(a^3+abc)$, here $x_1=bcd/a^3$ and $y_1=bc/a^2$. Expressing $y$'s via $x$'s is a linear system of equations with circulant matrix. But, say, already for $n=5$ lambdas are equal to $28/41,-2/41,-5/41,7/41,0$, and such $c$ does not exist.

  • $\begingroup$ Very interesting approach! I think it works. As we can divide all the $\alpha_i$ wlog by a common factor, there will always such a c, and it will even be positive, so the sign function isn't needed. I think the last "=" after Jensen must be a "$\le$" since you use $\beta_k\le1$ don't you? And why do you mention $\min_c \sum_i |\alpha_i+c|\leq 1$? max? $\endgroup$ – Wolfgang Jan 6 '16 at 8:06
  • $\begingroup$ Last = is = since $\sum \beta_k=1$ $\endgroup$ – Fedor Petrov Jan 6 '16 at 9:27
  • $\begingroup$ OK now I see. The product over i yields the $\sum \beta_k$. Nice! $\endgroup$ – Wolfgang Jan 6 '16 at 9:50
  • $\begingroup$ maximum of $F(c):=\sum |\alpha_i+c|$ is of course infinite, so if minimum does not exceed 1, there exists $c$ such that $F(c)=1$ as we need. $\endgroup$ – Fedor Petrov Jan 6 '16 at 9:52
  • $\begingroup$ oh sure, sorry... I misread as $\min_i |\alpha_i+c|\leq 1$... $\endgroup$ – Wolfgang Jan 6 '16 at 10:16

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