algebraic geometry - If $p$ is a positive multivariate polynomial, does $1/p$ have polynomial growth?

I wanted to ask a separate question to focus on an elementary issue from my question Does the inverse of a polynomial matrix have polynomial growth?.

Let $p : \mathbb{R}^n \to \mathbb{R}$ be a polynomial in $n$ real variables, with real coefficients, and suppose that $p > 0$ on all of $\mathbb{R}^n$.

Does $1/p$ have (at most) polynomial growth? That is, are there constants $C,N$ such that $1/p(\mathbf{x}) \le C (1+|\mathbf{x}|)^N$ for all $\mathbf{x} \in \mathbb{R}^n$?

Of course this is true when $n=1$ because in that case we have $\lim_{x \to \pm \infty} p(x) = +\infty$, and so $1/p$ is actually bounded. In higher dimensions this is more subtle: consider for example $p(x,y) = x^2 + (1-xy)^2$ which is strictly positive for all $x,y$, yet $1/p$ is unbounded: consider $p(1/y,y)$ as $y \to \infty$. For this example, I can use some calculus to show that $1/p(\mathbf{x}) \le 1+\|\mathbf{x}\|^2 \le (1+\|\mathbf{x}\|)^2$, so $1/p$ does have polynomial growth. But I don't know what to do in general.

Answer

Yes. By Stengle's Positivstellensatz, we can find polynomials $f_1$ and $f_2$, such that each of $f_1$ and $f_2$ can be written as sums of squares of polyomials, and $p f_1 = 1+f_2$. (In Wikipedia's language, take $F = \emptyset$ and $W = \mathbb{R}^n$.) Then $1/p = f_1/(1+f_2) \leq f_1$, which is a polynomial.