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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.

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  • $\begingroup$ I presume you can't mimic one of the usual inductive proofs of the inequality, extracting manifestly positive quantities along the way? $\endgroup$
    – Steven Stadnicki
    Commented Aug 30, 2017 at 18:05
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    $\begingroup$ 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. $\endgroup$
    – Ofir Gorodetsky
    Commented Aug 30, 2017 at 20:15
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    $\begingroup$ You may find this old post of mine here interesting. $\endgroup$
    – Andrés E. Caicedo
    Commented Aug 30, 2017 at 21:00
  • $\begingroup$ 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. $\endgroup$
    – Alexey Ustinov
    Commented Jan 5, 2018 at 13:25

3 Answers 3

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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.

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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.

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There is also the following way. $$\sum_{cyc}(a^5-abcde)=\frac{1}{48}\sum_{sym}(2a^5-2abcde)=$$ $$=\frac{1}{48}\sum_{sym}(a^5-a^4b-ab^4+b^5+a^4b-a^3b^2-a^2b^3+ab^4+a^3b^2-2a^3bc+a^3c^2)+$$ $$+\frac{1}{48}\sum_{sym}(a^3bc-2a^2b^2c+b^3ac+a^2b^2c-2a^2bcd+a^2d^2c+a^2bcd-2abcd+e^2bcd)=$$ $$=\tfrac{1}{48}\sum_{sym}\left((a-b)^2((a^2+b^2+ab)(a+b)+e^3+abc+cde)+a^2c(b-d)^2\right)$$

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  • $\begingroup$ could you add some more explanations, like how the constant $\frac1{48}$ came in the first step? $\endgroup$
    – vidyarthi
    Commented Oct 2, 2019 at 20:20
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    $\begingroup$ @vidyarthi I used $\sum\limits_{sym}a^5=\frac{5!}{5}\sum\limits_{cyc}a^5=24\sum\limits_{cyc}a^5.$ $\endgroup$
    – Michael Rozenberg
    Commented Oct 2, 2019 at 21:00
  • $\begingroup$ thanks! So that seems something like the relation between cyclic and symmetric groups. By the way, could you refer a proper repository of such formulae and factorizations, that is a work dealing with symmetric and cyclic inequalities $\endgroup$
    – vidyarthi
    Commented Oct 2, 2019 at 21:33
  • $\begingroup$ @vidyarthi I do it every time again. $\endgroup$
    – Michael Rozenberg
    Commented Oct 3, 2019 at 3:09
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    $\begingroup$ For example $\sum\limits_{sym}a^2b=\frac{1}{6}\sum\limits_{cyc}a^2(b+c+d+e)$ because in the expression $\sum\limits_{cyc}a^2(b+c+d+e)$ we have $20$ terms and $\frac{5!}{20}=6.$ This thinking is very useful for the work with symmetric polynomials! $\endgroup$
    – Michael Rozenberg
    Commented Oct 3, 2019 at 11:05

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