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Ji Chen posted the following on AoPS:

Prove that there exist four polynomials $p_{1}, p_{2}, p_{3}, p_{4}$ in $x, y, z$ so that $$\left ( x^{2}+ y^{2}+ z^{2} \right )^{3}- 8\left ( z^{3}x^{3}+ x^{3}y^{3}+ y^{3}z^{3} \right )= p_{1}^{2}+ p_{2}^{2}+ p_{3}^{2}+ p_{4}^{2}$$

I like Ji Chen's sum of squares (SOS) decomposition, which is very hard to find. Here is my way of thinking about this problem.

If $f = ab + c = -ad + e$ with $b, c, d, e \geq 0 \implies f \geq 0$, we have a new $f$ is an SOS, $$f:=\dfrac{cd+ e}{b+ d}$$

I think finding SOS like this is very funny. One day, you try substitutions and you succeed, however, another day, that trick is not useful anymore. Four years ago, I found this formula. Very amazing result, I have thought of it till now, that's a travelled task.

Rodrigo de Azevedo
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    I know that brute force has been used by Ji : it probably came from this article. This is useful, of course, for existence, but doesn't give the most beautiful expansions. – Sarvesh Ravichandran Iyer Mar 02 '21 at 05:06
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    @haidangel You may give details for "$f= ab+ c= -ad+ e\Rightarrow f:=\frac{eb+ cd}{b+ d}$,": For example, to prove $f \ge 0$, we find $P, Q, M, N \ge 0$ and $k$ such that \begin{align} f &= k P + Q \ f &= -k M + N. \end{align} Then $f = \frac{QM + PN}{P+M} \ge 0$. – River Li Mar 03 '21 at 03:05
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    @haidangel Perhaps it is very difficult. And you need to give details as you can. – River Li Mar 03 '21 at 03:48
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    this can be done when the degree of the homogeneous polynomial is twice an odd number $n.$ The polynomial may be written as $ V^T G V,$ where $V$ is a column vector made of all the terms of degree $n$ in some order. If the original polynomial is semipositive, it may be that the matrix is positive semidefinite. I need to think about that some more. In any even degree, the thing can be written as such a quadratic form. – Will Jagy Mar 04 '21 at 04:35
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    you should be reading https://www.researchgate.net/publication/240268385_Sums_of_Squares_of_Real_Polynomials and later articles/books that cite it. – Will Jagy Mar 04 '21 at 16:39
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    http://www.mit.edu/~parrilo/pubs/files/PeyrlParrilo-ComputingSumOfSquaresDecompositionsWithRationalCoefficients.pdf – Will Jagy Mar 04 '21 at 16:49
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    IN YOUR: $f= ab+ c= -ad+ e\Rightarrow f:=\frac{eb+ cd}{b+ d},$ What do you mean by the letters $a,b,c,d,e,f ; ? ? ? $ – Will Jagy Mar 04 '21 at 18:06
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    @TeresaLisbon it seems the aops answer is based on the technique in the article I linked byPeyrl and Parrilo: there is an understandable example in section 2.2.2, Example 1. I still have no idea what haidangel wants. – Will Jagy Mar 04 '21 at 18:10
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    @haidangel I will read the article for my own good, thank you for bringing it to my attention. Actually haidangel approached me over this problem, unfortunately I am unable to be of service due to the lack of clarity, and to be fair, after seeing the AOPS answer I have to see if my eyeballs are still intact! – Sarvesh Ravichandran Iyer Mar 04 '21 at 18:13
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    @TeresaLisbon I found a lecture on the Motzkin polynomial $x^4 y^2 + x^2 y^4 - 3 x^2 y^2 + 1$ which is not a sum of squares of polynomials, but multiply it by $x^2 + y^2 + 1$ and the new polynomial is the sum of squares (of polynomials). Evidently one may find these squares by the Gram matrix technique, which is mostly about the material of http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr – Will Jagy Mar 04 '21 at 18:36
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    @WillJagy Very nice at a glance, I have always admired your contributions. I will look through those in more detail when I can. Thank you for the consideration of sending me these links, I do like reading through new material when I can. – Sarvesh Ravichandran Iyer Mar 04 '21 at 18:39
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    @WillJagy$,\quad f$ is an SOS. –  Mar 05 '21 at 01:12
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    @haidangel what books or articles or lecture notes do you have on this topic? Titles, please. – Will Jagy Mar 05 '21 at 01:16
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    @WillJagy, I have just one formula$,\quad f= ab+ c= -ad+ e\Rightarrow f:=\frac{eb+ cd}{b+ d},$ now$,\quad f$ is just two squares $eb$ and $cd$ as you see. –  Mar 05 '21 at 01:35
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    For example, in the link$,\quad\left ( c- a \right )^{2}- \left ( a- b \right )\left ( b- c \right )= \left ( c+ a- 2b \right )^{2}+ 3\left ( a- b \right )\left ( b- c \right ):=H,$ we'll have $H=\frac{\left ( c+ a- 2b \right )^{2}+ 3\left ( c- a \right )^{2}}{1+ 3},$ we have new $H$ as an SOS. –  Mar 05 '21 at 01:41
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    @haidangel In $f= ab+ c= -ad+ e\Rightarrow f:=\frac{eb+ cd}{b+ d}$, you should use e.g. $P, Q, M, N, k$ (see my comment before). An easy example: We want to prove that $f = a^2 + b^2 + c^2 - ab - bc - ca \ge 0$ for all real numbers $a, b, c$. To this end, we write $f = kP + Q$ and $f = -kM + N$, where $k = (a-b)(b-c)$, $P = 3$, $Q = (c + a - 2b)^2$, $M = 1$, $N = (c-a)^2$. Then from $Mf = kPM + QM$ and $Pf = -kMP + NP$, we have $(M+P)f = QM + NP$ which results in $f = \frac{QM + NP}{M + P} = \frac{(c+a-2b)^2 + 3(c-a)^2}{1 + 3} \ge 0$. – River Li Mar 05 '21 at 03:12
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    So your question is asking for sum of squares of this polynomial in the left hand side? – NKellira Mar 06 '21 at 23:57
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    @haidangel OK. You need to give more details about $f= ab+ c= -ad+ e\Rightarrow f:=\frac{eb+ cd}{b+ d}$. For example, explain the idea and give an easy example. – River Li Mar 07 '21 at 02:39
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    @haidangel I think that it is not clear enough. Someone may ask what is $a, b, c, d, e, f$. Also, I suggest you say "$f$ can be written as smth" rather than "a new $f$" since they are the same $f$. I suggest you first give the idea of DRIVE!SOS, then give an easy example. – River Li Mar 08 '21 at 10:33

1 Answers1

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The four polynomials $p_i(x,y,z)$ should be a combination of the following terms: \begin{align} \left\{xyz,yx^2,y^2x,zx^2,z^2x,zy^2,z^2y,x^3,y^3,z^3\right\} \end{align} other terms would create the different left hand side. I have tried to use brutal force with the help of the deterministic global optimizer BARON starting with 40 unknowns coefficients by minimizing the objective of the form \begin{align} ((a_{003}^2+b_{003}^2+d_{003}^2+c_{003}^2)-1)^2+ ((c_{030}^2+b_{030}^2+d_{030}^2+a_{030}^2)-1)^2+\dots \nonumber \end{align}
(where $a_{ijk},b_{ijk},c_{ijk},d_{ijk}$ are coefficients to polynomials $p_1,p_2,p_3,p_4$). The calculation recovered the following form of the polynomials $p_1,p_2,p_3,p_4$: \begin{align} p_1(x,y,z)&=a_1(yz(y+z)-y^3-z^3) + a_2x^2(y+z)-a_3x(y^2+z^2) \\ p_2(x,y,z)&=b_1yz(z-y)+b_2x^2(y-z)+b_3x(z^2-y^2) \\ p_3(x,y,z)&=c_1yz(y-z)+c_2x^2(y-z)+c_3x(z^2-y^2)+c_4(y^3-z^3) \\ p_4(x,y,z)&=d_1xyz+d_2x(y^2+z^2)-x^3+d_3(y^3+z^3-yz(y+z)) \end{align} where \begin{align} a_1&=0.44778749472139178777752 \\ a_2&=1.6556891275393716966846 \\ a_3&=1.2079562282512847914973 \\ b_1&=-1.9014217828335169269138 \\ b_2&=-0.55113503404381225525555 \\ b_3&=-0.5511281347996555002311 \\ c_1&=1.1890222040768996247806 \\ c_2&=0.75843958935644073537929 \\ c_3&=1.3269533603627028384153 \\ c_4&=0.56886478882077529117822 \\ d_1&=1.8623260525794982367387 \\ d_2&=0.31014344808027582978127 \\ d_3&=0.68984000724248439873065 \end{align} with error of the objective function $0.43$E-$014$. Hope it helps further. However, analytical expressions of these coefficients should be recovered to check if this solution is really correct.

Vítězslav Štembera
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    It is nice (+1). I think the coefficients are not important (since they are some algebraic numbers), the form is important. – River Li Mar 12 '21 at 01:50
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    I am afraid I do not understand the problem math.stackexchange.com/q/4031531/822157 well, can you reformulate it please? (what is given and what is the question) – Vítězslav Štembera Mar 12 '21 at 03:34