Do these surfaces intersect? 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.
 A: 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 $y-2v_j$ (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.
