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Prove that there exist four polynomials $p_{1}, p_{2}, p_{3}, p_{4}$ in $x, y, z$ so that $$\left ( x^{2}+ y^{2}+ z^{2} \right )^{3}- 8\left ( z^{3}x^{3}+ x^{3}y^{3}+ y^{3}z^{3} \right )= p_{1}^{2}+ p_{2}^{2}+ p_{3}^{2}+ p_{4}^{2}$$ Source: AoPS/@Ji_Chen_ on.AoPS

I like Ji Chen's sum of squares, which is very hard to make decomposition like they said. Here is the way of thinking of mine about this problem_

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

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. I found this formula_ on.StackMath 4 years ago. Very amazing result, I have thought of it till now, that's a travelled task.

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    $\begingroup$ 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. $\endgroup$ – Teresa Lisbon Mar 2 at 5:06
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    $\begingroup$ @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 $f = \frac{QM + PN}{P+M} \ge 0$. $\endgroup$ – River Li Mar 3 at 3:05
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    $\begingroup$ @haidangel Perhaps it is very difficult. And you need to give details as you can. $\endgroup$ – River Li Mar 3 at 3:48
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    $\begingroup$ 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 \; ? ? ? $ $\endgroup$ – Will Jagy Mar 4 at 18:06
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    $\begingroup$ @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. $\endgroup$ – Will Jagy Mar 4 at 18:10
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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.

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    $\begingroup$ It is nice (+1). I think the coefficients are not important (since they are some algebraic numbers), the form is important. $\endgroup$ – River Li Mar 12 at 1:50
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    $\begingroup$ 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) $\endgroup$ – Vítězslav Štembera Mar 12 at 3:34
  • $\begingroup$ Okay, I'll try. $\endgroup$ – haidangel Mar 12 at 3:48

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