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 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.  

$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{6}\left ( \sum\limits_{cyc}\,(\,-\,3\,yz+ 3\,xy+ y^{\,2}+ z^{\,2}- 2\,x^{\,2}\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{2}\left ( \sum\limits_{cyc}\,(\,-\,yz- zx+ 2\,xy+ z^{\,2}- x^{\,2}\,)^{\,2} \right )$$
After that, using $ab+ bc+ ca= 0\!\therefore\!a\equiv (a- b)(b- c),b\equiv (b- c)(c- a),c\equiv (c- a)(a- b)$, so
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,x^{\,2}+ 2\,y^{\,2}- 3\,z^{\,2}- 4\,xy- yz+ 5\,zx\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,2\,x^{\,2}+ y^{\,2}- 3\,z^{\,2}- 5\,xy+ yz+ 4\,zx\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,3\,x^{\,2}- y^{\,2}- 2\,z^{\,2}- 5\,xy+ 4\,yz+ zx\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{14}\left ( \sum\limits_{cyc}\,(\,3\,y^{\,2}- 2\,z^{\,2}- x^{\,2}- xy- 4\,yz+ 5\,zx\,)^{\,2} \right )$$
because we always have the following equality with $\sum\limits_{cyc}\left ( (\,a- b\,)(\,b- c\,) \right )\left ( (\,b- c\,)(\,c- a\,) \right )= 0$
$$\left ( \sum\limits_{cyc}\left ( (\,a- b\,)(\,b- c\,) \right )^{\,2} \right )(\,x^{\,2}+ y^{\,2}+ z^{\,2}\,)= \sum\limits_{cyc}\,\left ( \sum\limits_{cyc}\,\left ( x(\,a- b\,)(\,b- c\,) \right ) \right )^{\,2}$$

Furthermore, we have $\sum\limits_{cyc}\,(\,-\,3\,yz+ 3\,xy+ y^{\,2}+ z^{\,2}- 2\,x^{\,2}\,)= 0= \left ( \sum\limits_{cyc}(\,a- b\,) \right )$. To apply
$$\sum\limits_{cyc}\,\frac{(\!a- b\!)^{2}}{2}\!=\!\frac{(\!a+ b- 2c\!)^{2}\!+\!3(\!a- b\!)^{2}}{4}\!=\!\frac{(\!b+ c- 2a\!)^{2}\!+\!3(\!b- c\!)^{2}}{4}\!=\!\frac{(\!c+ a- 2b\!)^{2}\!+\!3(\!c- a\!)^{2}}{4}$$
$$(\!\sum\limits_{cyc}\,x^{\,2}\!)^{\,2}- 3(\!\sum\limits_{cyc}\,x^{\,3}y\!)= \frac{(\!2\,x^{\,2}- y^{\,2}- z^{\,2}- 3\,xy+ 3\,yz\!)^{\,2}+ 3(\!y^{\,2}- z^{\,2}- xy- yz+ 2\,zx\!)^{\,2}}{4}$$
