Reduce quadratic form of $3$ variables to sum of 3 squares

Question

I am dealing with following quadratic form: $q:\mathbb{R}^3\rightarrow \mathbb{R}$ $$ q(x,y,z)=xy+yz+xz $$ and I am trying to reduce it to sum of squares. I approach this by trying to eliminate one variable. Since it is all symmetric, let's say we eliminate $x$. I proceed: $$ q=x(y+z)+yz=\bigg(x^2+x(y+z)+\bigg(\frac{y+z}{2}\bigg)^2 \bigg)-x^2-\bigg(\frac{y+z}{2}\bigg)^2+yz$$ which ... doesn't work, since there's stil all of $x,y,z$. Simply, for any variable $a$ there is no two terms: $a^2$ and $ab$ (for some other variable $b$) so i am not able to simplify it to a square this way. How to approach this?

Answer 1

As Ben Grossmann said, $q$ is not a sum of squares, but it can always be written as $\sum_i \alpha_i \varphi_i^2$ where $\alpha_i\neq 0$ and $\varphi_i$ is a linear form.

Nevertheless, you are in the nasty case (the case where there is no squared terms). The trick is to choose two variables (say $x$ and $y$), isolate all the terms containing at least one of the two variables , write it under the form $A [xy+B(z)x+C(z)y]$, where $A$ is a nonzero constant, and $B,C$ are terms in $z$ only, then "complete the product", and use the identity $uv=\dfrac{1}{4}((u+v)^2-(u-v)^2)$.

For you example $q(x,y,z)= 1. xy + z.x+z.y= (x+z)(y+z)-z^2=\dfrac{1}{4}(x+y+2z)^2-\dfrac{1}{4}(x-y)^2-z^2$.

This approach may be generalized to an arbitrary number of variables.