TITLE: Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms
QUESTION [9 upvotes]: The quartic form 
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$ 
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $Q_1$, $Q_2$, $Q_3$ so that
$$(x^2 + y^2 + z^2)^2 - 3( x^3 y + y^3 z + z^3 x) = Q_1^2 + Q_2^2 + Q_3^2$$
I would like to find an explicit writing of the quartic forms, with rational quadratic forms $Q_i$. Maybe more than $3$ terms are necessary.

REPLY [4 votes]: Using Macaulay2 (version 1.16) with package SumsOfSquares:
i1 : needsPackage( "SumsOfSquares" );

i2 : R = QQ[x,y,z];

i3 : f = (x^2 + y^2 + z^2)^2 - 3*( x^3 * y + y^3 * z + z^3 * x);

i4 : sosPoly solveSOS f

           1 2         1      1      1 2 2    9    1 2   2 2               1 2 2
o4 = (3)(- -x  + x*y - -x*z - -y*z + -z )  + (-)(- -x  + -y  + x*z - y*z - -z )
           2           2      2      2        4    3     3                 3

o4 : SOSPoly

Printing:
i5 : tex o4

o5 = $\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{3,\,\frac{9}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{-\frac{1}{2}\,x^{2}+x\,y-\frac{1}{2}\,x\,z
     -\frac{1}{2}\,y\,z+\frac{1}{2}\,z^{2},\,-\frac{1}{3}\,x^{2}+\frac{2}{3}\,y^{2}+x\,z-y\,z-\frac{1}{3}\,z^{2}\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$

In $\TeX$:
$$\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{3,\,\frac{9}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{-\frac{1}{2}\,x^{2}+x\,y-\frac{1}{2}\,x\,z
 -\frac{1}{2}\,y\,z+\frac{1}{2}\,z^{2},\,-\frac{1}{3}\,x^{2}+\frac{2}{3}\,y^{2}+x\,z-y\,z-\frac{1}{3}\,z^{2}\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$$
Hence,
$$\boxed{ f = 3 \left( -\frac{1}{2}\,x^{2}+x\,y-\frac{1}{2}\,x\,z
 -\frac{1}{2}\,y\,z+\frac{1}{2}\,z^{2} \right)^2 + \frac{9}{4} \left( -\frac{1}{3}\,x^{2}+\frac{2}{3}\,y^{2}+x\,z-y\,z-\frac{1}{3}\,z^{2} \right)^2 }$$

polynomials sum-of-squares-method macaulay2