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I have heard (if I am not mistaken) that there exists the following conjecture (or theorem?).

Let $u_1,\dots,u_n$ be unit vectors in an $n$-dimensional Euclidean vector space. Then there exists another unit vector $x$ such that $$(\prod_{i=1}^n |(x,u_i)|)^{1/n}\geq 1/\sqrt{n}.$$

Is it conjecture or theorem? In either case I would be interested to have a reference.

Remark. This post is a continuation of the previous one: Reference to a conjecture on unit vectors in Euclidean space

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    $\begingroup$ Yep, this one (to the best of my knowledge) is still open for large $n$ (small $n$ are fine, say, for $n\le 5$, the same idea as I used for the sum works and you can modify it slightly to go up a little bit. It also works for a small perturbation of an orthonormal system, so the counterexample, if it exists, should be rather skewed). Unfortunately, I'm not good with references, so I'll leave that part to someone else. $\endgroup$
    – fedja
    Commented Feb 22, 2020 at 20:18

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Not an answer. I tried the log sum inequality applied to the log of the left hand side, but no luck.

Assume the unit vectors $u_i$ are orthogonal and use a change of basis so that $u_i=e_i,$ the vector with the sole nozero entry equal to one in the $i^{th}$ component.

Taking $$x=\frac{1}{\sqrt{n}}(\pm 1,\pm 1,\ldots,\pm 1),$$ achieves your lower bound and perturbing $x$ only increases the LHS of your inequality.

Now consider the $u_i$ below with only two vectors changed and $\theta \in (0,1)$:

$$ \begin{array}{ccccccc} (& \sqrt{1-\theta^2} & \theta & 0 & \cdots & 0& ),\\ (& \theta &\sqrt{1-\theta^2} & 0 & \cdots & 0& ),\\ (& \sqrt{1-\theta^2} & \theta & 0 & \cdots & 0& ),\\ &&& \ddots &&&\\ (&0 & 0& 0 &\cdots & 1&)\\ \end{array} $$ and note that once again the left hand side of the inequality increases.

I feel that arguments along this line may prove the inequality.

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    $\begingroup$ this was too long to enter as a comment. I'd be interested to see if the downvoter has something constructive to contribute to the question or my comment. $\endgroup$
    – kodlu
    Commented Feb 19, 2020 at 22:08
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    $\begingroup$ I didn't downvote but I'm confused why you're bringing up whether the downvote can contribute to the question. If they have a reason to downvote your answer, that is enough; they don't need to be able to contribute to the question $\endgroup$
    – mathworker21
    Commented Feb 22, 2020 at 23:48
  • $\begingroup$ For orthogonal vectors the inequality is satisfied by $ x = \frac{y}{||y||}$, with $y = \sum_i u_i$. $\endgroup$
    – sigmatau
    Commented Feb 26, 2020 at 18:25

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