Writing $1-xyzw$ as a sum of squares

Question

Can you write $1 - xyzw$ in the form $p + q (1 - x^{2}-y^{2}-z^{2}-w^{2})$ where $p$ and $q$ are polynomials that are of the form $\sum g_{i}^{2}$ where $g_{i}$ $\in$ $\mathbb{R}[x,y,z,w]$? For instance, in the two variable case, $1 - xy = \frac{1}{2} + \frac{1}{2}(x-y)^{2} + \frac{1}{2}(1-x^{2}- y^{2})$. In this example, $q = \frac{1}{2}$, so we have a very simple expression. I'm looking for an analogous expression in four variables ($p$ and $q$ here can be anything as long as they are sums of squares). Also can we say anything in general for $2n$ variables?

Answer 1

The question would be even more interesting if the factor $1-x^2-y^2-w^2-z^2$ was replaced by $4-x^2-y^2-w^2-z^2$. The reason is that a necessary condition for the existence of such $p,q$ is that whenever $xywz>1$, the last factor is also negative. And it turns out that $$(xywz>1)\Longrightarrow(x^2+y^2+w^2+z^2<4),$$ by AM-GM inequality.

Notice that if the answer is positive with this constant $4$, then it is so with your constant $1$.

Now the answer: $p$ and $q$ exist for the sharp constant $4$. You may take $q=\frac1{16}(4+x^2+y^2+w^2+z^2)$ and $$p=\frac1{16}(x^2+y^2+w^2+z^2)^2-xywz.$$ To see $p$ is a sum of squares, just notice $$48p=(x^2+y^2-w^2-z^2)^2+\cdots+4(x-y)^2(w+z)^2+\cdots,$$ where the first sum consists of $3$ terms, and the second one consists in $6$ terms.

Answer 2

Take $q=x^2+y^2+z^2+w^2=:s$, then $$p=1-xyzw+s^2-s=\frac12(s-1)^2+\frac12+\frac12s^2-xyzw=\frac12(s-1)^2+\frac12+\frac12(xy-zw)^2+\dots$$ is a sum of squares.