# 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.

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.

• I presume you can't mimic one of the usual inductive proofs of the inequality, extracting manifestly positive quantities along the way? – Steven Stadnicki Aug 30 '17 at 18:05
• The even $n$ case is already due to A. Hurwitz (1891). The fact that a positivity certificate (in a slightly more general form then you require) exists in this case follows from Hilbert's 17th problem, solved by Artin (1927). The problem (now theorem) states: A polynomial assuming only non-negative values is a sum of squares of rational functions. This special case is of course simpler and the squares are of polynomials. – Ofir Gorodetsky Aug 30 '17 at 20:15
• You may find this old post of mine here interesting. – Andrés E. Caicedo Aug 30 '17 at 21:00
• You can constract such Certificate step by step if you'll follow usual proof of Muirhead's inequality en.wikipedia.org/wiki/Muirhead%27s_inequality On each step you just drop down one box at Young diagram. For examle replacing $x^n+y^n$ by $x^{n-1}y+y^{n-1}x$ (times smth) you'll get a positive summand $(x^{n-1}-y^{n-1})(x-y)$ (times smth) in your Certificate. – Alexey Ustinov Jan 5 '18 at 13:25

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.
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*}