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!
$\newcommand{\R}{\mathbb{R}}$ 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.