Wanted: Positivity certificate for the AM-GM inequality in low dimension I'm seeking for a Certificate of Positivity for 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 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 homogeneous of degree $1$ and positive (with arguments $\ge0\,$),
  
*each $\,s_i\,$ is a square?
  

In the case of $3$ variables the answer would be yes by the common factorisation
$$a^3+b^3+c^3-3abc\;=\;\frac 12(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right].$$
This is a Cross-post
from math.SE after a decent period of waiting ...
Remark: From David's comment to
this post
the $n=5$ expression does not factor according to Maple, contrary to the preceding $\,n=3\,$ case.

Added in edit:
I am really delighted by the community's rich spectrum of reactions, such a Math Overflow within the 12 hours after posting!
Thanks a lot!
In particular I've gotten a more general answer than hoped for, covering the specific issue addressed.
If you'd like to see a specific five-variables-certificate as initially sought-after, then you may follow the above "Cross-post" link, where a corresponding answer has been added.
 A: Let $$\phi(x_1,\cdots,x_n)=\frac{x_1^n+\cdots+x_n^n}{n}-x_1x_2\cdots x_n$$
In his proof of AG inequality, Hurwitz (1891) proves that
$$\phi(x_1,\cdots,x_n)=\frac{1}{2\times n!}\left(\phi_1+\phi_2+\cdots+\phi_n\right)$$
where
\begin{align*}
  \phi_1=& \sum(x_1^{n-2}+x_1^{n-3}x_2+\cdots+x_1x_2^{n-3}+x_2^{n-2})(x_1-x_2)^2,\\
\phi_2=&\sum(x_1^{n-3}+x_1^{n-4}x_2+\cdots+x_1x_2^{n-4}+x_2^{n-3})(x_1-x_2)^2x_3,\\
\dots\\
\phi_n=&\sum(x_1-x_2)^2x_3x_4\cdots x_n.
\end{align*}
Hurwitz, A. (1891). Ueber den Vergleich des arithmetischen und des geometrischen Mittels. Journal für die reine und angewandte Mathematik, 108, 266-268.
A: The following paper:

Fujiwara, Kazumasa, and Tohru Ozawa. Identities for the Difference between the Arithmetic and Geometric Means, (2014). 

proves the following representation for odd $n$:
\begin{equation*}
\frac{1}{n}\sum_i x_i^n - \prod_i x_i = \sum_{i=1}^n x_i\sum_{j \in J(n)} (P_{ij}(x_1,\ldots,x_n))^2,
\end{equation*}
for suitable polynomials $P_{ij}$. For even $n$, a SOS representation is available in Ch.2 of Hardy, Littlewood, Polyá.
A: There is also the following way.
$$\sum_{cyc}(a^5-abcde)=\frac{1}{48}\sum_{sym}(2a^5-2abcde)=$$
$$=\frac{1}{48}\sum_{sym}(a^5-a^4b-ab^4+b^5+a^4b-a^3b^2-a^2b^3+ab^4+a^3b^2-2a^3bc+a^3c^2)+$$
$$+\frac{1}{48}\sum_{sym}(a^3bc-2a^2b^2c+b^3ac+a^2b^2c-2a^2bcd+a^2d^2c+a^2bcd-2abcd+e^2bcd)=$$
$$=\tfrac{1}{48}\sum_{sym}\left((a-b)^2((a^2+b^2+ab)(a+b)+e^3+abc+cde)+a^2c(b-d)^2\right)$$
