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From Representing as sum of squares of polynomials I know that not every nonnegative polynomial is a sum of squares of polynomials. A classic counterexample is the Motzkin polynomial $$ M(x,y) = x^4 y^2 + x^2 y^4 + 1 - 3 x^2 y^2, $$ However, $M$ does admit a clean SOS decomposition if we allow cube roots of polynomials. Setting $$ u=((x^2y)^{1/3})^2, \quad v=((xy^2)^{1/3})^2, \quad w=1, $$$$ M(x,y) = u^3+v^3+w^3 - 3uvw = \tfrac12 (u+v+w)\big[(u-v)^2+(v-w)^2+(w-u)^2\big], $$ So here’s my question:

Is there a 3-variable nonnegative polynomial that still resists any SOS representation even if cube-root expressions are allowed?

Or could it be that any nonnegative polynomial can be written as a sum of squares built from polynomials and cube roots?

I know that with rational functions allowed (Artin’s solution to Hilbert’s 17th problem), every nonnegative polynomial works, but I am interested in this stricter “cube roots only” setting.

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  • $\begingroup$ The positivity of the Motzkin polynomial is based on the AM-GM inequality in the homogeneous case (which makes the appearance of cube roots here make sense). If you generalize this construction, I'd expect that other roots may be needed (but I don't have an example). Overall, I doubt that your original conjecture is true, but I don't have examples to show it. See, for example, this paper. $\endgroup$ Commented Sep 4 at 14:10

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