For a polynomial $f \in \mathbb{R}[x\_1, \cdots, x\_n]$, we say that $f$ is *coercive* (see my earlier question: [Real polynomials that go to infinity in all directions: how fast do they grow?](https://mathoverflow.net/questions/444925/real-polynomials-that-go-to-infinity-in-all-directions-how-fast-do-they-grow)) if


$$\displaystyle \lim\_{\lVert \mathbf{x} \rVert} f(\mathbf{x}) = \infty.$$


More precisely, $f$ is coercive if for all $c \in \mathbb{R}$ there exists $M(c)$ such that whenever $\lVert \mathbf{x} \rVert \geq M(c)$ we have $f(\mathbf{x}) \geq c$. Further, it is known that there is always a positive *order of coercivity* $q(f)$ satisfying the property that for all $c > 0$ there exists $M(c)$ satisfying


$$\displaystyle \lVert \mathbf{x} \rVert \geq M(c) \Rightarrow f(\mathbf{x}) \geq c \lVert \mathbf{x} \rVert^{q(f)}.$$


It is known that $q(f)$ can be arbitrarily small (see: [How fast do coercive polynomials grow?](http://num.math.uni-goettingen.de/preprints/files/2017-1.pdf)).


My question is: does the following statement hold?


Let $f \in \mathbb{R}[x\_1, \cdots, x\_n]$ be coercive. Then there exists a positive number $c\_1$ and a real number $c\_2$ such that


$$\displaystyle f(x\_1, \cdots, x\_n) \geq c\_1 \min\{|x\_1|, \cdots, |x\_n|\}^2 - c\_2.$$



Lower bound for coercive polynomials
A counterexample is given by
$$
f(x,y)=(x^4-y^3)^2+y .
$$
Clearly, $f(R, R^{4/3})=R^{4/3}\ll R^2 = \left( \min \{ R, R^{4/3} \} \right)^2$, so $f$ doesn't satisfy your condition.


However, $f$ is coercive: If $y\le 0$, then $f(x,y)\ge x^8+y^6 +y$ and now either $|y|$ is not large and there are no problems or if $|y|\gg 1$, then $y^6+y\ge (1/2)y^6$, say.


If $y\ge 0$, then $f(x,y)\ge \max\{ (x^4-y^3)^2, y\} \to\infty$ as $|(x,y)|\to\infty$, $y\ge 0$.


