Suppose there are $k$ non-zero homogeneous polynomials $P^1(.),\cdots, P^k(.)$, each of degree $r$ in $n$ variables, such that $P^j(x_1,\cdots,x_n) \geq 0$ for all $(x_1,\cdots,x_n) \geq 0$, for all $j \in [k]$. Under what conditions (on the $P^j(.)$'s) would there exist an $\alpha \in \mathbb{R}^n_{+}$ such that $P^j(\alpha) > 0$ for all $j \in [k]$?
I can see that for each $j$ there exists an $\alpha^j \in \mathbb{R}^n_{+}$ such that $P^j(\alpha^j) > 0$ (otherwise $P^j(.)$ would become the zero polynomial). But not sure whether and how we can construct an $\alpha$ that will work for all $P^j(.)$'s.
EDIT: I tried to solve the above question as follows.
Fix $j$. Suppose there exists an open ball $B_{\epsilon}(\overline{x}) \subset \mathbb{R}^n_{+}$ such that $P^j(x) = 0$ for all $x \in B_{\epsilon}(\overline{x})$. But this means $P^j(x) = 0$ for all $x \in \mathbb{R}^n_{+}$. (Is this true? I'm not able to prove this as of now but it seems true) Hence $P^j(x) = 0$ for all $x \in \mathbb{R}^n_{+}$. Hence $P^j(.)$ is the zero polynomial. Contradiction. Therefore for each $j$, the set $\{x \in \mathbb{R}^n_{+}: P^j(x)=0\}$ is a measure zero set in $\mathbb{R}^n_{+}$. Hence the union of such sets for all $j$ is also measure zero. Hence for almost all $x \in \mathbb{R}^n_{+}$ $P^j(x) > 0$ for all $j$.
But I'm not able to prove exactly how being equal to zero for an open ball of points implies being the zero polynomial.
