Wanted: Low-dimensional SOS certificate for the AM-GM inequality Consider the AM-GM inequality in five variables
$$a^5+b^5+c^5+d^5+e^5-5abcde\;\ge 0\qquad\forall\,a,b,c,d,e\ge 0\,.$$

Can one write the LHS as a concrete (finite) sum
  $\,\sum_i h_i\,s_i\,$ with real polynomials
  $\,h_i(a,b,c,d,e)\,$ and $\,s_i(a,b,c,d,e)$, where each
  $\,h_i\,$ is positive and homogeneous of degree $1$, and each $\,s_i\,$ is a square?

For the AM-GM inequality in $3$ variables the answer would be yes, cf
$$a^3+b^3+c^3-3abc\;=\;\frac 12(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$$
which is a common factorisation of the LHS.
If $\,n\,$ is even, we'd expect the $\,h_i\,$ to be homogeneous of degree $0\,$, thus reducing to positive scalars. Corresponding instances in $4$ and $8$ variables are given here:
$$\begin{eqnarray}a^4+b^4+c^4+d^4-4abcd = &\;\left(a^2-b^2\right)^2
+\left(c^2-d^2\right)^2 + 2(ab-cd)^2\\[2ex]
a^8+b^8+c^8+d^8+e^8+f^8+g^8+h^8-8abcdefgh = &\:
\left(a^4-b^4\right)^2 + \left(c^4-d^4\right)^2\qquad\\[1ex]
&  + \left(e^4-f^4\right)^2 + \left(g^4-h^4\right)^2\\[1ex]
& + 2\,\left((ab)^2-(cd)^2\right)^2 + 2\,\left((ef)^2-(gh)^2\right)^2 \\[1ex]
& + 4\,\left(abcd - efgh\right)^2.\end{eqnarray}$$
All these Sum Of Squares (SOS) positivity certificates share the special properties as asked for in the question. And for $\,n=5\,$
(to which "low-dimensional" in the title refers) I'm curious to see one!
My question is a follow-up to this post from which two notes can be drawn:


*

*In M. Rozenberg's answer you'll find an SOS expression for $\,n=5\,$, but without the features as formulated above – at least not obviously?!

*From David's comment: According to Maple the $n=5$ expression cannot be written as a product, contrary to the $\,n=3\,$ expression above.

 A: Yes, one can.
An SOS certificate of positivity having the desired properties is
$$\begin{eqnarray}
a^5+b^5+c^5+d^5+e^5-5abcde\;= & \;\frac14\sum_\text{cyclic}a\left[\left(b^2-c^2\right)^2+\left(d^2-e^2\right)^2+2(bc-de)^2\right] \\[.5ex]
&+\frac1{12}\sum_\text{cycl}\,a\,(a-b)^2 \left(2a^2 +(a+b)^2\right) \\[.5ex]
&+\frac 1{12}\sum_\text{cycl}\,a\,(a-c)^2 \left(2a^2 +(a+c)^2\right)\\[.5ex]
&+\frac 1{12}\sum_\text{cycl}\,a\,(a-d)^2 \left(2a^2 +(a+d)^2\right)\\[.5ex]
&+\frac 1{12}\sum_\text{cycl}\,a\,(a-e)^2 \left(2a^2 +(a+e)^2\right),
\end{eqnarray}$$
it has been produced by K. Fujiwara and T. Ozawa.
The corresponding paper is referenced to in this MO answer, obtained after cross-posting.
To check the preceding expression one may use Python's CAS module sympy:
>>> from sympy import *
>>> a, b, c, d, e = symbols('a b c d e')
>>> simplify(
(a*((b**2-c**2)**2 +(d**2-e**2)**2 +2*(b*c-d*e)**2)
+b*((c**2-d**2)**2 +(e**2-a**2)**2 +2*(c*d-e*a)**2)
+c*((d**2-e**2)**2 +(a**2-b**2)**2 +2*(d*e-a*b)**2)
+d*((e**2-a**2)**2 +(b**2-c**2)**2 +2*(e*a-b*c)**2)
+e*((a**2-b**2)**2 +(c**2-d**2)**2 +2*(a*b-c*d)**2))/4
+(
  a*((a-b)**2 *(2*a**2 +(a+b)**2) +(a-c)**2 *(2*a**2 +(a+c)**2) +(a-d)**2 *(2*a**2 +(a+d)**2) +(a-e)**2 *(2*a**2 +(a+e)**2))
+ b*((b-a)**2 *(2*b**2 +(b+a)**2) +(b-c)**2 *(2*b**2 +(b+c)**2) +(b-d)**2 *(2*b**2 +(b+d)**2) +(b-e)**2 *(2*b**2 +(b+e)**2))
+ c*((c-a)**2 *(2*c**2 +(c+a)**2) +(c-b)**2 *(2*c**2 +(c+b)**2) +(c-d)**2 *(2*c**2 +(c+d)**2) +(c-e)**2 *(2*c**2 +(c+e)**2))
+ d*((d-a)**2 *(2*d**2 +(d+a)**2) +(d-b)**2 *(2*d**2 +(d+b)**2) +(d-c)**2 *(2*d**2 +(d+c)**2) +(d-e)**2 *(2*d**2 +(d+e)**2))
+ e*((e-a)**2 *(2*e**2 +(e+a)**2) +(e-b)**2 *(2*e**2 +(e+b)**2) +(e-c)**2 *(2*e**2 +(e+c)**2) +(e-d)**2 *(2*e**2 +(e+d)**2))
)/12 )

yields
a**5 - 5*a*b*c*d*e + b**5 + c**5 + d**5 + e**5

