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We have the following fact: (I don't remember where I read it, but there is.)

If $f(x)$ is which is non-negative for $x\ge 0,$ then $f(x)=g(x)+x\cdot h(x),$ where $g(x)$ and $h(x)$ are SOS.

So I wanna know how to find $f(x,y,z)$ as $$f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z),$$ where $g(x,y,z),p(x,y,z)$ are all non-negative polynomials.

That also means that my problem is:

Given $f(x,y,z),$ which is a non-negative polynomial where $x,y,z\ge 0.$ Find $g(x,y,z)$ and $p(x,y,z)\ge 0$ (for same condition) such as $f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z).$

There would be something like @Haidangel's question for make

$$H:= m\left ( a, b, c \right )+ kn\left ( a, b, c \right )ca;$$ and $$H:=m\!\left(a,b,c\right)\!-n\!\left(a,b,c\right)\!\left(a-b\right)\!\left(b-c\right)$$ where $m(a,b,c),n(a,b,c)$ are non-negative.

But here, I wanna make it as a sum of polynomials.

I tried $$f(x,y,z)=f(0,y,z)+\left[f(x,y,z)-f(0,y,z)\right]\equiv f(0,y,z)+x\cdot h(x,y,z),$$

But it's not the form I like since $h(x,y,z)$ is not certainly as SOS.

Also, AoPS/@dragonheart6 had told me to use The Unknown coefficients but I don't know how to do it here because there are so many coefficients for the higher degree problems.

Finally, my last purpose is to make $f(x,y,z)$ as SOS where the condition of variables is $x,y,z$ are all non-negative.

$\textbf{Note.}$ We know that if replace $x,y,z$ by $a^2,b^2,c^2$ then we have to prove the inequality for real numbers, but now, the degree is very high.

I wish that someone could have a solution and an example about this.

PS. SOS means Sum of Squares.

NKellira
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    I got curious about sos; the articles written about it bear no resemblance to what haidangel writes as questions. If a polynomial really is a sum of squares, there is an algorithm, the first part was called the Gram matrix method; there are programs to do exctly this. Next, if a polynomial $g$ is positive semidefinite, usually there is a factor $h = (x_1^2 + x_2^2 + \cdots x_n^2)^r$ in the same $n$ variables such that $gh$ is a sum of polynomials squared. Precise conditions are in a result by Reznick. Anyway, haidangel talks about doing research, but seems not to be reading books or articles – Will Jagy Mar 07 '21 at 00:48
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    @WillJagy There is a SOStools program to do SOS for real numbers. But it's not the generality, only some inequality can give the exact SOS, it can not give SOS with sqrt sign. Also, it can't give SOS for the non-negative numbers, example: https://artofproblemsolving.com/community/u493456h1846414p20593473 – NKellira Mar 07 '21 at 00:52
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    “We all know the following fact” – well, I know it, you know it, and probably many know it, but all? It is always good to add a reference for such a statement. – Martin R Mar 10 '21 at 09:26
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    The rest of the question is unclear to me: Find $g(x,y,z)$ and $p(x,y,z)$ such that $f(x,y,z):=x\cdot g(x,y,z)+p(x,y,z)$ satisfied what? – Martin R Mar 10 '21 at 09:27
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    @MartinR I mean that: Given $f(x,y,z)$ is a non-negative polynomials where $x,y,z\ge 0.$ Find $g(x,y,z)$ and $p(x,y,z)\ge 0$ such as $f(x,y,z)=x\cdot g(x,y,z)+p(x,y,z).$ – NKellira Mar 10 '21 at 09:45
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    Then I suggest to write that in your question, instead of the vague “I wanna know how the way you usually use to make $f$ ...” – Martin R Mar 10 '21 at 09:48
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    @MartinR I will edit soon. – NKellira Mar 10 '21 at 09:49
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    Are you asking if such polynomials exist (for given $f$) or how to compute them? – Martin R Mar 10 '21 at 09:50
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    @MartinR I want to compute them, I'm editing to make it clearly. – NKellira Mar 10 '21 at 09:51
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    For polynomials that are non-negative on interval: See theorem 5 in https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-972-algebraic-techniques-and-semidefinite-optimization-spring-2006/lecture-notes/lecture_10.pdf. OR first page of https://www.ams.org/journals/tran/2000-352-10/S0002-9947-00-02595-2/S0002-9947-00-02595-2.pdf – River Li Mar 10 '21 at 15:23
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    I think you may say something like "If $f(x)$ is a polynomials which is non-negative for " rather than "If $f(x)$ is nonnegative-polynomials for". – River Li Mar 10 '21 at 15:26
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    @RiverLi Yeah, I edited, thanks. – NKellira Mar 11 '21 at 03:41

1 Answers1

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We have the following fact: (I don't remember where I read it, but there is.)

If $f(x)$ is which is non-negative for $x\ge 0,$ then $f(x)=g(x)+x\cdot h(x),$ where $g(x)$ and $h(x)$ are SOS.

This result can be found in the book by Pólya and Szegö (1st ed, p. 78): enter image description here For higher dimensions it is not so easy. You are trying to use Putinar's Positivstellensatz, meaning you are looking for SOS polynomials $g_1$, $g_2$, $g_3$ and $p$ such that: $$f(x,y,z)=x\cdot g_1(x,y,z)+y\cdot g_2(x,y,z)+z\cdot g_3(x,y,z)+p(x,y,z).$$

The conditions for that theorem are not satisfied for nonnegativity over $\mathbb{R}^n_+$, so not every nonnegative $f$ will admit such a representation with $g_i$ and $p$.

The Krivine–Stengle Positivstellensatz states that every polynomial $f$ that satisfies $f(x)\geq 0$ for all $x\in\mathbb{R}^3_+$ can be decomposed using SOS polynomials $g_1,\ldots,g_6$ as: $$f(x)\left(x_1\cdot g_4(x)+x_2\cdot g_5(x)+x_3\cdot g_6(x)\right) = f(x)^{2k} + x_1\cdot g_1(x)+x_2\cdot g_2(x)+x_3\cdot g_3(x)$$ for some $k\in\mathbb{N}$. The classic method to find $g_1,\ldots,g_6$ is using semidefinite optimization (which was already mentioned in the comments). YALMIP is another excellent tool for that. This method requires you to have an upper bound on the degree of the polynomials, which you do not have. You also do not know $k$. What you could do is vary $k$ and gradually increase the upper bound and hope you get 'lucky'. I am not aware of any other general method.

A nice background read is Section 3 (3.3 and 3.6 in particular) of Sums of squares, moment matrices and optimization over polynomials by M. Laurent.

LinAlg
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  • Very nice. After reading pages from numerous articles, then the book Squares by Rajwade, I still do not know SOS expressions of the two Choi and Lam (about 1975) examples (immediately nonnegative by AM-GM ) page 101 in Rajwade, first appearance pages 75,76. the 4,4 is $w^4 +x^2y^2 +y^2z^2 + z^2 x^2 - 4wxyz$ and the 3,6 is $x^4y^2 +y^4z^2 + z^4 x^2 - 3x^2y^2z^2$ Lots of related material by Reznick https://faculty.math.illinois.edu/~reznick/ I think the original Choi and Lam paper was 1976 – Will Jagy Mar 16 '21 at 17:03
  • and Delzell's 1997 example comes from the Choi and Lam sextic in $x,y,z,$ multiply by a new $w^2$ but add an eighth power of an original, $z^8 + w^2 x^4 y^2 + w^2 y^4 z^2 + w^2 z^4 x^2 - 3 w^2 x^2 y^2 z^2 ; . ; ;$ This example, multiplied by any $(w^2 + x^2 + y^2 + z^2)^k$ is never sos of polynomials, but multiplying it by $(x^2 + y^2 + z^2)$ does give sos http://www.math.ens.fr/%7Ebenoist/textes/Delzell.pdf – Will Jagy Mar 16 '21 at 17:33
  • @WillJagy you probably know this, so just for those others who are confused: not every nonnegative polynomial is SOS, see Thm 3.4 in the background read by Laurent. – LinAlg Mar 16 '21 at 17:37
  • Right. The motivation of this OP seems to lead to lengthy discussions on the aops site, primarily looking for rapid ways to solve problems, perhaps rapid ways to prove apparent conclusions. The Laurent article looks good. Seems I was unclear in my first comment, I suspect that the two Choi-Lam examples can be taken into SOS simply by multiplying by the appropriate $(w^2 + x^2 + y^2 + z^2)$ – Will Jagy Mar 16 '21 at 17:47
  • But Yalmip can’t give exact SOS, huh? – NKellira Mar 16 '21 at 23:21
  • @WillJagy 1st one: $(y^2+z^2)(w^4-4wxyz+x^2y^2+x^2z^2+y^2z^2)$ is an SOS. SOStools (in Matlab) can give exact SOS. You can also first simplify it to $w^4-4wxyz+x^2y^2+x^2z^2+y^2z^2 = \left( {y}^{2}+{z}^{2} \right) \left( x-2,{\frac {wyz}{{y}^{2}+{z}^ {2}}} \right) ^{2}+{\frac {{w}^{4}{y}^{2}+{w}^{4}{z}^{2}-4,{w}^{2}{y} ^{2}{z}^{2}+{y}^{4}{z}^{2}+{y}^{2}{z}^{4}}{{y}^{2}+{z}^{2}}} $, then find SOS for ${w}^{4}{y}^{2}+{w}^{4}{z}^{2}-4,{w}^{2}{y}^{2}{z}^{2}+{y}^{4}{z}^{2}+ {y}^{2}{z}^{4} $. – River Li Mar 17 '21 at 00:37
  • @WillJagy 2nd one: $ \left( {x}^{2}+{y}^{2}+{z}^{2} \right) \left( {x}^{4}{y}^{2}-3,{x}^ {2}{y}^{2}{z}^{2}+{z}^{4}{x}^{2}+{y}^{4}{z}^{2} \right) $ is an SOS. – River Li Mar 17 '21 at 00:41
  • @RiverLi I asked a separate question on the early Choi-Lam examples, https://math.stackexchange.com/questions/4064630/choi-lam-homogeneous-polynomials-as-sums-of-squares Willie Wong found the first one by hand. Oh: I don't have matlab. – Will Jagy Mar 17 '21 at 00:42
  • @WillJagy $F = z^8 + w^2 x^4 y^2 + w^2 y^4 z^2 + w^2 z^4 x^2 - 3 w^2 x^2 y^2 z^2$ is interesting since $(x^2 + y^2 + z^2 + w^2)^k F$ is not SOS. Hmm, it is $F = \left( {x}^{4}{y}^{2}-3,{x}^{2}{y}^{2}{z}^{2}+{z}^{4}{x}^{2}+{y}^{4} {z}^{2} \right) {w}^{2}+{z}^{8} $. – River Li Mar 17 '21 at 00:52
  • @RiverLi right, that is Delzell's best known example. The point is that $w^2$ times the Choi-Lam sextic is not the sum of squares, as the sextic is not. Anyway, I threw that one in at the end of my new question – Will Jagy Mar 17 '21 at 00:59
  • Is there any proof or reference for statement 45. I don't have access to the linked book at the moment. – Kvothe May 08 '21 at 13:34
  • @Kvothe the proof is in the same figure... – LinAlg May 08 '21 at 13:48
  • @LinAlg, are you saying that the 45 below the line gives the proof? Where does the first statement (so not the identity to be applied) come from. Would you mind explaining the proof in a little more detail? Perhaps in my question https://math.stackexchange.com/questions/4131009/sums-of-squares-and-positivity-on-the-half-line-related-to-hilbert-theorem-188? – Kvothe May 09 '21 at 15:49
  • Also I guess there is typo. Should $x +x_1$ be equal to something? $x_1$ is never even defined. – Kvothe May 09 '21 at 16:19