# How to find parameters under which a six-order polynomial in four variables is globally nonnegative?

Let

$$f(\xi_1,\xi_2,\xi_3,\xi_4)=\frac{1}{120} \xi _3 \xi _1^5\left(15 (\alpha -6) (\alpha-5) (\alpha -4) (\alpha-2)+120 c_3 (\alpha +\beta-5)\right)+c_1 \xi _2^2 \xi_1^4 (\alpha +\beta-5)+\frac{1}{6} \xi _2 \xi _3\xi _1^3 \left(6 (\alpha -4) (\alpha -2) (2 \alpha -9)+6c_2 (\alpha +\beta -4)+12c_1+24c_3\right)+\frac{1}{12} \xi_2^3 \xi _1^2 \left(12 c_4(\alpha +\beta -4)+36c_1\right)+\frac{1}{4} \xi_3^2 \xi _1^2 \left(2 (\alpha-2) (7 \alpha -24)+4 c_5 (\alpha +\beta -3)+4c_2\right)+\frac{1}{2} \xi_2^2 \xi _3 \xi _1 \left(3(\alpha -2) (3 \alpha -11)+2c_6 (\alpha +\beta -3)+4c_2+6 c_4\right)+\xi _3 \xi_4 \xi _1 \left(2 \alpha-\beta +2c_5-4\right)+\frac{1}{2} \xi_2 \xi _3^2 \left(10 (\alpha-2)+2 c_5+4 c_6\right)+c_3\xi _4 \xi _1^4+c_2 \xi _2\xi _4 \xi _1^2+c_4 \xi_2^4+c_6 \xi _2^2 \xi _4-\xi_4^2$$

be a polynomial of degree $$6$$ with $$\alpha,\beta>0$$ and $$c_1,c_2,c_3,c_4,c_5,c_6\in\mathbb{R}$$. My question is as follows:

Do there exist $$\alpha,\beta>0$$ and $$c_1,c_2,c_3,c_4,c_5,c_6\in\mathbb{R}$$ such that $$f(\xi_1,\xi_2,\xi_3,\xi_4)\le 0$$ for all $$\xi_1,\xi_2,\xi_3,\xi_4\in\mathbb{R}$$?

I'd like to write Mathematica codes to solve this problem, but I have no idea to start with it.

Any reference, suggestion, idea, or comment is welcome. Thank you!

• According to the Abel-Ruffini theorem, the general equation of degree n for n≥5 is unsolvable in radicals. It is likely that this circumstance is "pulling" and problems when working with non-polynomial inequalities, especially with so many additional parameters.
– dtn
Feb 6, 2020 at 7:23
• Either a term analysis of the polynomial can help here, based on the inadmissibility of non-fulfillment of inequality requirements (if possible). Perhaps there are methods for reducing a polynomial of degree 6 to a polynomial of a lower order and its subsequent analysis. I would start with this. Let's see what other users prompt. <en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem>
– dtn
Feb 6, 2020 at 7:23
• What if we only want a solution, i.e. find one set of solution $(\alpha,\beta,c_1,c_2,c_3,c_4,c_5,c_6)$? It occurs to me that perhaps the command "FindInstance" works?
– LCH
Feb 6, 2020 at 7:54

I think you are right in choosing which function to use FindInstance.

But, I'm not sure that the problem can be solved analytically using this command, therefore, for a start I set the parameters of $$\xi_i$$ equal to some (random) numbers.

pars = {ξ1 = 1, ξ2 = 1, ξ3 = 1, ξ4 = 1}

eqn = 1/120 ξ3 ξ1^2 (15 (α -
6) (α - 5) (α -
4) (α - 2) +
120 c3 (α + β - 5)) +
c1 ξ2^2 ξ1^4 (α + β - 5) +
1/6 ξ2 ξ3 ξ1^3 (6 (α -
4) (α - 2) (2 α - 9) +
6 c2 (α + β - 4) + 12 c1 + 24 c3) +
1/12 ξ2^3 ξ1^2 (12 c4 (α + β - 4) +
36 c1) +
1/4 ξ3^2 ξ1^2 (2 (α - 2) (7 α -
24) + 4 c5 (α + β - 3) + 4 c2) +
1/2 ξ2^2 ξ3 ξ1 (3 (α -
2) (3 α - 11) +
2 c6 (α + β - 3) + 4 c2 +
6 c4) + ξ3 ξ4 ξ1 (2 α - β +
2 c5 - 4) +
1/2 ξ2 ξ3^2 (10 (α - 2) + 2 c5 + 4 c6) +
c3 ξ4 ξ1^4 + c2 ξ2 ξ4 ξ1^2 + c4 ξ2^4 +
c6 ξ2^2 ξ4 - ξ4^2

FindInstance[α > 0 && β >
0 && {c1, c2, c3, c4, c5, c6} ∈ Rationals &&
eqn <= 0, {α, β, c1, c2, c3, c4, c5, c6}]


And here is an example of the results of using this function, under which the given conditions are satisfied.

{{α -> 1, β -> 1, c1 -> 0, c2 -> 0, c3 -> 0, c4 -> 0,
c5 -> 0, c6 -> 0}}


You can introduce additional conditions in the form of $$\xi_i \epsilon R$$ and include these variables as the desired coefficients. Thus, you will get all the information about the value of the parameters of the equation for which the requirements of the inequality are satisfied.

FindInstance[α > 0 && β >
0 && {c1, c2, c3, c4, c5, c6} ∈ Rationals && {ξ1, ξ2, ξ3, ξ4} ∈ Rationals &&
eqn <= 0, {α, β, c1, c2, c3, c4, c5, c6, ξ1, ξ2, ξ3, ξ4}]

{{α -> 1, β -> 1, c1 -> 0, c2 -> 0, c3 -> 0, c4 -> 0,
c5 -> 0, c6 -> 0, ξ1 -> 0, ξ2 -> 0, ξ3 -> 0, ξ4 -> 0}}

• Thank! Now the question becomes, what if $\xi_1,\xi_2,\xi_3,\xi_4\in\mathbb{R}$ are arbitrary?
– LCH
Feb 6, 2020 at 10:07
• I corrected my answer as an answer to your question.
– dtn
Feb 6, 2020 at 10:34
• If the arbitrary of these parameters changes the mechanism for solving the problem, then these parameters are not set in Mathematica like pars. Then, the problem is not solved analytically.
– dtn
Feb 6, 2020 at 10:40
• Ah sorry, perhaps I didn't make it clear. I need that inequality to be true for all $\xi_1,\xi_2,\xi_3,\xi_4\in\mathbb{R}$. $\xi_1,\xi_2,\xi_3$, and $\xi_4$ are not viewed as parameters $\alpha, beta$, and $c_i$.
– LCH
Feb 6, 2020 at 14:37
• It's the first time for me to ask a question here. How can I paste my Mathematica codes here? I think of some possible codes to solve this problem.
– LCH
Feb 6, 2020 at 14:39

I figure out the following codes to solve:

Reduce[Exists[{α, β, c1, c2, c3, c4, c5, c6}, ForAll [{ξ1,ξ2,ξ3,ξ4}, eqn<= 0 && α>0 && β>0]], Reals]

However, it run for a long time and no result can be obtained.

• I guess it's because there are lots of parameters and variables. Any other possible way to solve this problem?
– LCH
Feb 6, 2020 at 16:48
• The calculation cannot end in any way, because an analytical solution is sought, and it must be enormous. I see here only a numerical solution to the problem.
– dtn
Feb 6, 2020 at 17:05
• A numerical solution is also ok. Could you give some hint how to get a numerical solution?
– LCH
Feb 6, 2020 at 17:18
• Please see my answer. My hint is this: you yourself set the parameters $\xi$ so that it belongs to a real numbers R. And then you get solutions for the desired parameters in the form of certain numbers. Having done this several times, you will accumulate a small "database", and then you can build the dependencies of these parameters from each other and approximately understand the nature of their change when changing the parameters of the set $\xi$. Perhaps this is all I can help here. I hope that I have brought you even closer to a solution.
– dtn
Feb 6, 2020 at 17:30