Motzkin's original proof shows that $x^4y^2 + x^2y^4 + z^6 - a x^2y^2z^2 x^2y^2z^2$ is psd and not sos for any a $a$ in the interval (0,3]. $(0,3]$. If you take $a = .02 02$ say, it is reasonably simple, though messy, to show that (x^4y^2 $(x^4y^2 + x^2y^4 + z^6 - .02x^2y^2z^2)^3 02x^2y^2z^2)^3$ is a sum of squares; in fact, it's a sum of binomial squares (x^b $(x^b y^c z^d - x^e y^f z^g)^2z^g)^2$, where b+c+d=e+f+g=9. $b+c+d=e+f+g=9$. The idea is to look at any monomial with a negative coefficient and make it into the middle term of this square, in a way that the other two terms are still in the Newton polytope. For example, one term in the given cube is -.06x^10y^6z^2, $-.06x^10y^6z^2$, which is "handled" by .03(x^6y^3 $.03(x^6y^3 - x^4y^3z^2)^2x^4y^3z^2)^2$. It's sort of messy to work out, but I've convinced myself (at least) that it's true.