Let $x,y,z,w$ be four variables over $\mathbb{R}$. Let $A,B,C \in \mathbb{R}[x,y,z,w]$ be three homogeneous polynomials, each of degree $3$.
Is there a real solution to: $B \leq A, B \leq C$? Namely, is there $s=(\alpha,\beta,\gamma,\delta) \in \mathbb{R}^4$, such that $B(s) \leq A(s), B(s) \leq C(s)$?
Remarks: (1) I am interested in the existence of such $s \in \mathbb{R}^4$, not in how to find it exactly. Moreover, I am asking for $s \in \mathbb{R}^4$, not just $s \in \mathbb{C}^4$.
(2) Maybe it is possible to obtain a solution by just taking $(x,y,z,w)=(1,1,1,w)$, if (at least) one of the three $\{A,B,C\}$ has degree $3$ in $w$, and work in $\mathbb{R}[w]$ (and apply this). However, I am not sure if this is a solution (probably not necessarily), since it may happen that this solves only one of the two inequalities, not both. (But remember that I have taken very specific $(x,y,z)$, namely, $(1,1,1)$).
(3) Perhaps this question is relevant.
I really apologize if my question is quite obscure; it is just that I have calculated something and arrived at such a situation. Thank you very much!
Let $\tilde{A}(s) = A(s) - B(s)$ and $\tilde{C}(s) = C(s) - B(s)$. Since $A, B, C$ are homogeneous of degree $3$, so do $\tilde{A}$ and $\tilde{C}$. For any $s \in \mathbb{R}^4$, we have $\tilde{A}(-s) = -\tilde{A}(s)$ and $\tilde{C}(-s) = -\tilde{C}(s)$.
Using continuity of $\tilde{A}$, we can find a point $t \in \mathbb{R}^4$ with $|t| = 1$ and $\tilde{A}(t) = \tilde{A}(-t) = 0$. Since at least one of $\tilde{C}(t)$ or $\tilde{C}(-t) \ge 0$. We have find a $s = t$ or $-t$ such that $|s| = 1$ and
$$\tilde{A}(s) = 0, \tilde{C}(s) \ge 0 \quad\iff\quad A(s) = B(s), C(s) \ge B(s)$$