Let $f(\xi_1,\xi_2,\xi_3,\xi_4):=a_2 \xi _2^4+a_1 \xi _1^2 \xi _2^3+a_4 \xi _1 \xi _3 \xi _2^2+a_8 \xi _4 \xi _2^2+a_6 \xi _3^2 \xi _2+a_3 \xi _1^3 \xi _3 \xi _2+a_7 \xi _1^2 \xi _4 \xi _2+a_5 \xi _1^2 \xi _3^2+a_9 \xi _1 \xi _3 \xi _4$ be a polynomial of order 5. Then what's the conditions on $a_i$ ($i=1,...9$) such that $f(\xi_1,\xi_2,\xi_3,\xi_4)\le 0$ for all $\xi_1,\xi_2,\xi_3,\xi_4\in\mathbb{R}$?
It seems to me that the coefficients $a_1$,$a_3$, $a_4$, $a_6$, $a_7$, $a_8$, and $a_9$ must vanish. Then only $a_2$ and $a_5$ are to be determined. I am not sure about this.
Any reference, suggestion, idea, or comment is welcome. Thank you!
(I also want to ask which fields do such problems belong to? Any appropriate references?)
I write $(x,y,z,w) = (\xi_1,\xi_2,\xi_3,\xi_4)$.
With $x=z=w=0$ we find $a_2y^4 \leqslant 0$ for all $y\in\Bbb R$, so $a_2\leqslant 0$.
With $w=z=0$, we find $a_1x^2y^3+a_2y^4 \leqslant 0$ for all $x,y\in\Bbb R$.
With $y=1$, we find $a_1x^2+a_2\leqslant 0$ for all $x\in\Bbb R$, and hence $a_1\leqslant 0$.
With $y=-1$, we find $-a_1x^2+a_2\leqslant 0$ for all $x\in\Bbb R$, and hence $a_1\geqslant 0$. We conclude $a_1 = 0$.
With $x=w=0$, we find $a_2y^4 + a_6yz^2\leqslant 0$ for all $y,z\in\Bbb R$.
With $y=1$ we find $a_2+a_6z^2\leqslant 0$ for all $z\in\Bbb R$, so $a_6\leqslant 0$.
With $y=-1$ we find $a_2-a_6z^2\leqslant 0$ for all $z\in\Bbb R$, so $a_6\geqslant 0$. We conclude $a_6 = 0$.
With $x=z=0$ and $y=1$, we find $a_8w+a_2\leqslant 0$ for all $w\in\Bbb R$, and so $a_8 = 0$.
With $y=0$ and $x=z=1$, we find $a_9w + a_5 \leqslant 0$ for all $w\in\Bbb R$, so $a_9 = 0$.
With $x = y=1, z=0$ we find $a_7w+a_2\leqslant 0$ for all $w\in\Bbb R$, so $a_7 = 0$.
With $y=z=1$ we get $a_3x^3+a_5x^2+a_4x + a_2 \leqslant 0$ for all $x\in\Bbb R$, so $a_3 = 0$ and $a_5\leqslant 0$.
This leaves us with
$$a_2y^4 + a_4xy^2z + a_5x^2z^2.$$
Set $C = a_2y^4\leqslant 0$, $B = a_4y^2$ and $t=xz$. We must hence have
$$a_5t^2+Bt + C\leqslant 0$$
for all $t\in\Bbb R$. For this to happen, the discriminant $B^2-4a_5C$ must be $\leqslant 0$, so
$$a_4^2y^4\leqslant 4a_5a_2y^4$$
must hold for all $y$. In other words, $|a_4|\leqslant 2\sqrt{a_2a_5}$.
It follows that all coefficients must be $0$ except for $a_2$, $a_4$ and $a_5$. We must have $a_2,a_5 \leqslant 0$ and $|a_4|\leqslant 2\sqrt{a_2a_5}$.