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For any $a_{1}, a_{2}, \dots, a_{6} \in \mathbb{R}$ with

$$\sum_{i=1}^{6}a_{i}^{2}=1$$

is it true that there always exist $x_{1}, x_{2}, \dots, x_{6} \in \mathbb{R}$ with $\displaystyle\sum_{i=1}^{6}x_{i}^{2}=6$ such that

$$x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}x_{5}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}=1? $$

Any helpful answer that could lead to a correct answer to this question would be highly appreciated!

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migrated from mathoverflow.net Apr 14 at 21:56

This question came from our site for professional mathematicians.

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    $\begingroup$ Let $f(x) = f(x_1, \ldots, x_6)$ be the value of the expression on the left side of the displayed equation. At one extreme, choose $x_i = \text{sign}(a_i) \in \{-1, 1\}$ for each $i$. Then surely $f(x) = M = \left(\sum_{i=1}^6 |a_i|\right)^2 \geq 1$ since $\sum_{i=1}^6 a_i^2 = 1$. At another extreme, $f(x) = 0$ as soon as $x_2 = 0$ say (e.g., $x = (\sqrt{6}, 0, \ldots, 0)$. Since $\{x \in \mathbb{R}^6| \sum_{i=1}^6 x_i^2 = 6\}$ is connected, its image under $f$ is connected and therefore contains every value in the interval $[0, M]$, including $1$ in particular. $\endgroup$ – user43208 Apr 14 at 13:22
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    $\begingroup$ Wasn't there a question very similar just a couple of days ago? $\endgroup$ – Lee David Chung Lin Apr 14 at 22:07
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$\newcommand{\R}{\mathbb{R}}$ The answer is yes. Indeed, without loss of generality $a_i\ge0$ for all $i$. Let \begin{equation} S:=\Big\{(x_1,\dots, x_6)\in\R^6\colon\sum_{i=1}^6x_i^2=6\Big\} \end{equation} and let the function $f\colon S\to\R$ be defined by \begin{equation} f(x_1,\dots, x_6):=(x_1\cdots x_6)^2\Big(\sum_{i=1}^6a_ix_i\Big)^2. \end{equation} Since $S$ is connected and $f$ is continuous, the set $f(S)$ is an interval in $\R$. Moreover, \begin{equation} 0=f(0,\dots,0,\sqrt6)\in f(S) \end{equation} and \begin{equation} f(1,\dots,1)=\Big(\sum_{i=1}^6a_i\Big)^2\ge\sum_{i=1}^6a_i^2=1. \end{equation} So, \begin{equation} 1\in\big[f(0,\dots,0,\sqrt6),f(1,\dots,1)\big]\subseteq f(S). \end{equation} That is, there is $(x_1,\dots, x_6)\in S$ such that $f(x_1,\dots, x_6)=1$, as desired.

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    $\begingroup$ I don't think your inequality is right: suppose the $a_i$'s cancel each other. Of course this is easily repaired, but maybe this question is too trivial for MO? $\endgroup$ – user43208 Apr 14 at 13:27
  • $\begingroup$ I see that, while I was typing this answer, Todd Trimble made a comment with essentially the same content. $\endgroup$ – Iosif Pinelis Apr 14 at 13:28
  • $\begingroup$ @ToddTrimble : I had assumed, without loss of generality, that $a_i\ge0$ for all $i$, which will prevent the cancellation. $\endgroup$ – Iosif Pinelis Apr 14 at 13:30
  • $\begingroup$ Oh you did; sorry. $\endgroup$ – user43208 Apr 14 at 13:32
  • $\begingroup$ This is the kind of argument that one learns in 1st year analysis courses. $\endgroup$ – Mere Scribe Apr 14 at 14:31

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