# On solution to the equation $x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}x_{5}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}=1$

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

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

## migrated from mathoverflow.netApr 14 at 21:56

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• 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. – user43208 Apr 14 at 13:22
• Wasn't there a question very similar just a couple of days ago? – Lee David Chung Lin Apr 14 at 22:07

$$\newcommand{\R}{\mathbb{R}}$$ The answer is yes. Indeed, without loss of generality $$a_i\ge0$$ for all $$i$$. Let $$$$S:=\Big\{(x_1,\dots, x_6)\in\R^6\colon\sum_{i=1}^6x_i^2=6\Big\}$$$$ and let the function $$f\colon S\to\R$$ be defined by $$$$f(x_1,\dots, x_6):=(x_1\cdots x_6)^2\Big(\sum_{i=1}^6a_ix_i\Big)^2.$$$$ Since $$S$$ is connected and $$f$$ is continuous, the set $$f(S)$$ is an interval in $$\R$$. Moreover, $$$$0=f(0,\dots,0,\sqrt6)\in f(S)$$$$ and $$$$f(1,\dots,1)=\Big(\sum_{i=1}^6a_i\Big)^2\ge\sum_{i=1}^6a_i^2=1.$$$$ So, $$$$1\in\big[f(0,\dots,0,\sqrt6),f(1,\dots,1)\big]\subseteq f(S).$$$$ That is, there is $$(x_1,\dots, x_6)\in S$$ such that $$f(x_1,\dots, x_6)=1$$, as desired.
• 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? – user43208 Apr 14 at 13:27
• @ToddTrimble : I had assumed, without loss of generality, that $a_i\ge0$ for all $i$, which will prevent the cancellation. – Iosif Pinelis Apr 14 at 13:30