SOS: Proof of the AM-GM inequality

Question

"A basic strategy in tackling inequalities of few variables is to write things into sum of squares..." is a quote from an answer, intended as commenting the post entitled "Can this inequality proof be demystified?".

I'd like to know what the Sum Of Squares strategy yields for the AM-GM inequality with $\,n$ variables. In the low-dimensional cases $\,n=2,3,4\,$ one has \begin{align*} a^2+b^2-2ab &\;=\;(a-b)^2\;\geqslant\;0\,, \\[2ex] a^3+b^3+c^3-3abc &\;=\;(a+b+c)\cdot\underbrace{\frac 12\left((a-b)^2+(b-c)^2+(c-a)^2\right)}_{a^2+b^2+c^2-ab-bc-ca}\;\geqslant\;0\,,\\[2ex] 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\;\geqslant\;0\,, \end{align*} and for $\,n=5\,$ cf this math.SE answer.

Now fix some $n\in\mathbb{N}$, let $\,a_1,a_2,\cdots,a_n\geqslant 0$, and consider $$\sum_{k=1}^n a_k^n\, -\, n\prod_{k=1}^n a_k\tag{AM-GM}$$ which, I'm pretty convinced, can be expressed involving Sums Of Squares and then recognised to be non-negative.

• But how does such an expression look like?
• Can it be derived in a (a sort of) uniform procedure?

There's the post Proofs of AM-GM inequality (tagged as 'big-list') with 14 answers currently, but none of them would answer this question.

I've considered to add the 'computer-algebra-systems' tag.

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Added in edit
Pedro Tamaroff introduced the central keyword Certificate of positivity in his comment which, amongst further insights, led me to rediscover the above $n=4$ expression.

$\exists$ recommended readings:
(1) An algorithm for sums of squares of real polynomials, J. Pure & Appl. Alg. 127 (1998), pp 99–104, is a nice paper by Victoria Powers and Thorsten Wörmann
(2) Computing sum of squares decompositions with rational coefficients
by Helfried Peyrl and Pablo A. Parrilo, Theor. Comp. Sci. 409 (2008), pp 269-281

Answer 1

Yes, we can write $x_1^n+x_2^n+...+x_n^n-nx_1x_2...x_n$ in the SOS form, but I think it's very ugly and it's not useful for big $n$.

We can make the following.

For $n=3$: $$a^3+b^3+c^3-3abc=\sum_{cyc}\left(a^3-\frac{1}{2}a^2b-\frac{1}{2}a^2c\right)+\sum_{cyc}\left(\frac{1}{2}a^2b+\frac{1}{2}a^2c-abc\right)=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b)+\frac{1}{2}\sum_{cyc}c(a-b)^2=\frac{1}{2}(a+b+c)\sum_{cyc}(a-b)^2.$$

For $n=4$: $$a^4+b^4+c^4+d^4-4abcd=\sum_{cyc}\left(a^4-\frac{1}{3}(a^3b+a^3c+a^3d)\right)+$$ $$+\frac{1}{6}\sum_{sym}(a^3b-a^2b^2)+\frac{1}{6}\sum_{sym}(a^2b^2-a^2bc)+\frac{1}{3}\sum_{cyc}(a^2bc+a^2cd+a^2bd-3abcd)=$$ $$=\frac{1}{6}\left(\sum_{sym}(a^4-a^3b)+\sum_{sym}(a^3b-a^2b^2)+\sum_{sym}(a^2b^2-a^2bc)+\sum_{sym}(a^2bc-abcd)\right)=$$ $$=\frac{1}{12}\left(\sum_{sym}(a^4-a^3b-ab^3+b^4)+\sum_{sym}(a^3b-2a^2b^2+ab^3)+\sum_{sym}(c^2a^2-2c^2ab+c^2b^2)+\sum_{sym}(a^2cd-2abcd+b^2cd)\right)=$$ $$=\frac{1}{12}\sum_{sym}(a-b)^2(a^2+b^2+c^2+2ab+cd)=$$ $$=\tfrac{(a-b)^2(2(a+b)^2+(c+d)^2)+(a-c)^2(2(a+c)^2+(b+d)^2)+(a-d)^2(2(a+d)^2+(b+c)^2)}{6}+$$ $$+\tfrac{(b-c)^2(2(b+c)^2+(a+d)^2)+(b-d)^2(2(b+d)^2+(a+c)^2)+(c-d)^2(2(c+d)^2+(a+b)^2)}{6}.$$ Here $$\sum\limits_{sym}a=6(a+b+c+d)$$ $$\sum\limits_{sym}a^2b^2=4(a^2b^2+a^2c^2+b^2c^2+a^2d^2+b^2d^2+c^2d^2),...$$ For $n=5$: $$a^5+b^5+c^5+d^5+e^5-5abcde=\frac{1}{24}\sum_{sym}(a^5-abcde)=$$ $$=\frac{1}{24}\left(\sum_{sym}(a^5-a^4b)+\sum_{sym}(a^4b-a^3b^2)+\sum_{sym}(a^3b^2-a^3bc)\right)+$$ $$+\frac{1}{24}\left(\sum_{cyc}(a^3bc-a^2b^2c)+\sum_{sym}(a^2b^2c-a^2bcd)+\sum_{sym}(a^2bcd-abcde)\right)=$$ $$=\frac{1}{48}\left(\sum_{sym}(a^5-a^4b-ab^4+b^5)+\sum_{sym}(a^4b-a^3b^2-a^2b^3+ab^4)+\sum_{sym}(c^3a^2-2c^3ab+c^3b^2)\right)+$$ $$+\frac{1}{48}\left(\sum_{cyc}(a^3bc-2a^2b^2c+b^3ac)+\sum_{sym}(a^2c^2d-2c^2abd+b^2c^2d)+\sum_{sym}(a^2cde-2abcde+b^2cde)\right)=$$ $$=\frac{1}{48}\sum_{sym}(a-b)^2(a^3+2a^2b+2ab^2+b^3+c^3+abc+c^2d+cde).$$ The rest is the same.