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

Question

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!

Answer 1

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}}

Answer 2

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.