TITLE: CSB inquality: is $\|x\|^2\|y\|^2 - \langle x,y \rangle^2$ a square in any obvious way?
QUESTION [16 upvotes]: Suppose $x=(x_1,x_2),y = (y_1,y_2) \in \mathbb{R}^2$. I noticed that
\begin{align*}
\|x\|^2 \|y\|^2 - \langle x,y \rangle^2 &=
x_1^2y_1^2 + x_1^2 y_2^2 + x_2^2 y_1^2 + x_2^2 y_2 ^2 - (x_1^2 y_1^2 + 2 x_1 y_1 x_2 y_2 + x_2^2 y_2^2) \\
&=(x_1 y_2)^2 - 2x_1 y_2 x_2 y_1 + (x_2 y_2)^2 \\
&=(x_1 y_2 - x_2 y_1)^2
\end{align*}
which proves the CSB inequality in dimension two. This begs the question:

If $x = (x_1,\ldots,x_n),y=(y_1,\ldots,y_n) \in \mathbb{R}^n$, is there a polynomial $p \in \mathbb{R}[x_1,\ldots,x_n;y_1,\ldots,y_n]$ such that $ \|x\|^2 \|y\|^2 - \langle x,y \rangle^2 = p^2$?

REPLY [5 votes]: When $n\ge3$, $\|x\|^2\|y\|^2 - \langle x,y\rangle^2$ is not the square of any polynomial $p(x,y)$. Keep all entries other than $x_1$ fixed and let
\begin{align*}
q(x_1) &= \|x\|^2\|y\|^2 - \langle x,y\rangle^2,\\
\Rightarrow q\,'(x_1) &= 2(x_1 \|y\|^2 - y_1\langle x,y\rangle)
\end{align*}
If $q$ is a squared polynomial, some zero of $q\,'$ must be a zero of $q$. However, when $x=(x_1,1,0,0,\ldots,0)$ and $y=(1,0,1,0,0,\ldots,0)$, we have $q\,'(x_1)=2x_1$ and $q(x_1)=2(x_1^2+1)-x_1^2$. So, the only zero of $q\,'$ is $x_1=0$, but $q(0)=2\neq0$. Therefore $q$ is not a squared polynomial.