For any real numbers $a_{1},a_{2},\cdots a_{6}$ and $b_{1},b_{2},\cdots b_{6}$ with $\sum_{i=1}^{6}a_{i}^{2}=1$ and $\sum_{i=1}^{6}b_{i}^{2}=1$, does the equation $$ x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}\left(\sum_{i=1}^{6}b_{i}x_{i}\right)^{2}=1 $$ always have a solution $x_{1},x_{2},\cdots x_{6}$ in $\mathbb{R}$ with $\sum_{i=1}^{6}x_{i}^{2}=6$? Thanks.

2$\begingroup$ what have you tried? what is your background? this may be suitable in math stack exchange if you can edit this mentioning what all you know $\endgroup$ – Praphulla Koushik Jan 22 at 14:11

2$\begingroup$ @PraphullaKoushik I have tried special cases and the answer is yes for the special cases; and on the other hand, I could not find any counterexample. Best regards. $\endgroup$ – mathers1 Jan 22 at 14:16

1$\begingroup$ which special cases have you tried? add that in the question.. what techniques you know about intersection of surfaces add that in the question $\endgroup$ – Praphulla Koushik Jan 22 at 14:22

1$\begingroup$ @PraphullaKoushik I tried the special cases of choosing some of them to be 0 's or 1's. I am not in the area of algebraic surface, but this question is about intersection of surfaces, which I believe the researchers in the area of surfaces or related can answer. Thanks. $\endgroup$ – mathers1 Jan 22 at 14:29

2$\begingroup$ There have been a relatively large number of edits in a short space of time. It would be preferable to work out before hand what one wants to write, and then stick with it $\endgroup$ – Yemon Choi Jan 23 at 2:53
The answer is "yes" though the argument is rather ad hoc and doesn't generalize to vectors in more general positions.
We have $6$ unit vectors $v_j$ out of which the first $4$ are pairwise orthogonal and want to show that there exists a vector of length $\sqrt 6$ such that $\prod_{j=1}^6 \langle x,v_j\rangle\ge 1$ (to get below $1$ is trivial). Consider all sums $y=\sum_{j=1}^6\varepsilon_j v_j$ where $\varepsilon_j=\pm 1$ and choose the one with the largest length. Replacing some $v_j$ with $v_j$, if necessary, we can assume WLOG that it is $y=\sum_j v_j$. Comparing $y$ with $y2v_j$ (one sign flip), we see that $\langle y,v_j\rangle\ge 1$ for all $j$. Unfortunately, $y$ is a bit long, but it cannot get the length greater than $4$ (the $4$ pairwise orthogonal vectors produce length $2$) and we have $$ \y\^2=\sum_j \langle y,v_j\rangle=:\sum_j (1+u_j), \quad 0\le u_j\le 3 $$ Reducing the length to $\sqrt 6$ means that we have to multiply $y$ by $\left(1+\frac 16\sum_j u_j\right)^{1/2}$, so it suffices to show that $$ \prod_j(1+u_j)\ge \left(1+\frac 16\sum_j u_j\right)^3 $$ i.e. $$ \prod_j(1+u_j)^{1/3}\ge 1+\frac 16\sum_j u_j. $$ However, on $[0,3]$, we have $(1+u)^{1/3}\ge 1+\frac u6$ (the LHS is concave, so it is enough to check the endpoints) and Bernoulli finishes the story.

$\begingroup$ Thank you for your answer. When you multiply $y$ by $\left(1+\frac 16\sum_j u_j\right)^{1/2}$, how would you have $\langle y,v_j\rangle\ge 1$ for all $j$? Thanks. $\endgroup$ – mathers1 Feb 10 at 10:26

$\begingroup$ @mathers1 I would not. I would just have the product $\prod_j \langle x,v_j\rangle$ equal to $\left(1+\frac 16\sum_j u_j\right)^{3}\prod_j(1+u_j)$ and the rest of the post is about showing that it is still $\ge 1$. $\endgroup$ – fedja Feb 10 at 11:01