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For any integers $i$ and $j$ such as $1\le i<j\le6$, let $x_{ij}$ be a nonnegative real number.

Is it true that, given the condition
$$\sum_{1\le i<j\le6}x_{ij}^2=1,$$ the sum $$\sum_{1\le i<j<k\le6}x_{ij}x_{ik}x_{jk}$$ is maximized when all the $x_{ij}$'s are the same?

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    $\begingroup$ There's strong numerical evidence that this is indeed the case. $\endgroup$
    – Aryeh Kontorovich
    Commented Sep 17 at 21:07
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    $\begingroup$ @AryehKontorovich : Yes, my numerical experiments, too, suggested that this is true. $\endgroup$
    – Iosif Pinelis
    Commented Sep 17 at 21:20
  • $\begingroup$ The Lagrange multipliers’ method gives a system of $15$ equations $$\cases{a_{13} a_{23}+ a_{14} a_{24}+ a_{15} a_{25}+ a_{16} a_{26}=\lambda a_{12}\\ \vdots\\a_{15} a_{16}+ a_{25} a_{26}+ a_{35} a_{36}+ a_{45} a_{46}=\lambda a_{56}}$$ $\endgroup$
    – Pietro Majer
    Commented Sep 17 at 21:38
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    $\begingroup$ @PietroMajer : I tried something like that, but this does not hold for the diagonal elements. $\endgroup$
    – Iosif Pinelis
    Commented Sep 17 at 21:48
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    $\begingroup$ @PietroMajer yes, since from the proof of the inequality we see that on eigenvalues level the only equality case is when all but one are equal to, say, $-\alpha$, then $A+\alpha I$ has rank one and the rest is easy $\endgroup$
    – Fedor Petrov
    Commented Sep 18 at 13:14

2 Answers 2

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Put $x_{ij}=x_{ji}$ and consider the symmetriс matrix $A=(x_{ij})$ with zeros on diagonal. Then $\operatorname{tr} A=0$, $\operatorname{tr} A^2=2\sum_{i<j} x_{ij}^2=2$ is fixed, denote it $2=30 \alpha^2$ and what should be maximized is nothing but $\operatorname{tr} A^3=6\sum_{i<j<k} x_{ij}x_{ik}x_{jk}$, the conjectured maximal value is $6{6\choose 3}\alpha^3=120\alpha^3$. So, for eigenvalues $t_i$ we have $\sum t_i=0$, $\sum t_i^2=30\alpha^2$ and want to prove $\sum t_i^3\leqslant 120\alpha^3$. If $t_1=\ldots=t_5=-\alpha$ and $t_6=5\alpha$, we have equality (and this is realized when all $x_{ij}$ are equal to $\alpha$).

Well, let's try the following standard trick. Imagine that we have $(t_i-5\alpha)(t_i+\alpha)^2\leqslant 0$ for all $i$, then sum up and get the bound of $\sum t_i^3$ via $\sum t_i^2$ which turns into equality in our case, so this bound is $120\alpha^3$ (a direct check: what we get is $\sum t_i^3\leqslant 3\alpha\sum t_i^2+6\cdot 5\alpha^3=120\alpha^3$). Ok, if not, say, if $t_6>5\alpha$, then we have $t_1^2+\ldots+t_5^2\geqslant \frac{(t_1+\ldots+t_5)^2}5=\frac{t_6^2}5>5\alpha^2$, also $t_6^2>25\alpha^2$ and $\sum t_i^2>30\alpha^2$, a contradiction.

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This problem can be posed as finding a point satisfying a system of polynomial equations and inequalities. There exist a few software for solving them, such as QEPCAD (available in Sage) and RAGLib (available in Maple).

UPDATED. I've supplied the system composed of $\sum_{i<j} x_{i,j}^2=1$, $x_{i,j}\geq 0$, $c^2 = \binom{6}{3}^2/\binom{6}{2}^3$, $c>0$, and $\sum_{i<j<k} x_{i,j}x_{i,k}x_{j,k} > c$ to QEPCAD, requesting to find any point satisfying it, and it said that such point does not exist. However, along the way it comes with a lot of warnings about "The McCallum projection may not be valid." Here is this calculation at SageCell.

Maybe RAGlib can do a better job here.

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  • $\begingroup$ Hmm. What if your $c$ is negative? $\endgroup$
    – Iosif Pinelis
    Commented Sep 17 at 23:04
  • $\begingroup$ @IosifPinelis: Oh! I forgot to specify $c>0$, but that's for good, since absence of solution with negative $c$ is wrong. In verbose mode I see a lot of warnings like "WARNING! Projection factor ... is everywhere zero in the cylinder over the cell ... of postive dimension. The McCallum projection may not be valid.", which seems to affect the result. $\endgroup$
    – Max Alekseyev
    Commented Sep 17 at 23:20
  • $\begingroup$ Sorry, I don't know how to interpret this warning. Also, if my understanding is not mistaken, QEPCAD seems to have overlooked the case $c<0$, which makes its work seem questionable. $\endgroup$
    – Iosif Pinelis
    Commented Sep 18 at 1:15
  • $\begingroup$ @IosifPinelis: That was my concern as well. The result cannot be trusted because of those warnings. $\endgroup$
    – Max Alekseyev
    Commented Sep 18 at 1:32
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    $\begingroup$ Thank you for your answer anyway. I somehow had forgotten that this problem is one of real algebraic geometry, and your answer was a good reminder about that. Mathematica does not seem to be able to handle this problem. $\endgroup$
    – Iosif Pinelis
    Commented Sep 18 at 1:57

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