# 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.

• 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 – Praphulla Koushik Jan 22 at 14:11
• @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 counter-example. Best regards. – mathers1 Jan 22 at 14:16
• which special cases have you tried? add that in the question.. what techniques you know about intersection of surfaces add that in the question – Praphulla Koushik Jan 22 at 14:22
• @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. – mathers1 Jan 22 at 14:29
• 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 – Yemon Choi Jan 23 at 2:53

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.
• 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. – mathers1 Feb 10 at 10:26
• @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 $\ge 1$. – fedja Feb 10 at 11:01