You can decompose the region where $f \geq 0$ as a finite union of regions (with disjoint interiors) such that each region is the intersection of a finite number of sets, each set defined by an inequality $p(x_1, \dots, x_n) \geq 0$. A finite number of polynomials is enough to describe the locus where $f \geq 0$, but more than one polynomial could be necessary.
This is a consequence of Tarski's theorem on quantifier elimination for real-closed fields.
In the comments there was a request to solve the example $f(x,y,z) = \sqrt{x} + \sqrt{y} - z \geq 0$.
The domain is divided into four regions based on the sign of $x$ and $y$. In three of the regions $x$ or $y$ is negative and $f$ is undefined. The region where $f$ exists is the intersection of $x \geq 0$ and $y \geq 0$ and can be divided into two pieces, one with $z \leq 0$ and the second with $z \geq 0$. In the first piece $f \geq 0$ everywhere. In the second piece, $f \geq 0$ is equivalent to $z^2 \leq x + y + 2 \sqrt{xy}$ and one can subdivide again according to the sign of $q(x,y,z) = z^2 - x - y$. Where $q \leq 0$, then $f \geq 0$. Where $q \geq 0$, the condition is $4xy - q^2 \geq 0$.
The condition that $f$ is defined and non-negative can always be expressed as a decision tree in which each decision is a test of the sign of a polynomial.