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?) 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?).
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
$$\displaystyle f(x_1, \cdots, x_n) \geq c_1 \min\{|x_1|, \cdots, |x_n|\}^2 – c_2.$$
Best Answer
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 $|(x,y)|\to\infty$, $y\ge 0$.