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Let $G$ be the $6\times6$ Gram matrix of $6$ unit vectors in $\Bbb R^3$.

Can the mean of the squares of the off-diagonal entries of $G$ be $<1/5$?

Remark 1: A numerical experiment suggests that $1/5$ is the exact lower bound on the mean of the squares of the off-diagonal entries of $G$.

Remark 2: If the restriction "in $\Bbb R^3$" is removed, then, of course, one can make all the off-diagonal entries of $G$ zero.

Remark 3: If the $6$ unit vectors in $\Bbb R^3$ are chosen uniformly and independently at random, then the expectation of the square of each off-diagonal entry of $G$ is $1/3$, which is $>1/5$.

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    $\begingroup$ 1/5 is optimal, it is answered here: mathoverflow.net/a/273028/4312 $\endgroup$
    – Fedor Petrov
    Commented Sep 18 at 8:52
  • $\begingroup$ @FedorPetrov : Thank you for your comment. Can you expand it into a formal answer, so that I could accept it? $\endgroup$
    – Iosif Pinelis
    Commented Sep 18 at 9:04

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No, it can not. This is equivalent to the claim that the sum of squares of all elements of the Gram matrix is not less than $6+30/5=12=6^2/3$. This bound in wider generality ($n$ unit vectors in $\mathbb{R}^d$) is proved here

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