Decay of real continuous algebraic functions at infinity Let $f$ be a real valued continuous algebraic function on $\mathbb R^n$. Suppose the zero set of $f$ is bounded, i.e., if $|x|$ is large enough, $f(x)\neq 0$. Is there any estimate of the sort $|f(x)|\ge c(1+|x|)^{-N}$ for some $c>0$ and $n\in\mathbb N$? A prototypical example of such an inequality is $\sqrt{1+x^2}-x\ge c(1+|x|)^{-1}$, but I would appreciate general estimates along this line that apply to algebraic functions of $n$ variables.
 A: Update: As pointed out by Igor, this is not convincing. I only show that if $p(x,f)=0$ and $p(x,0)\not=0$ for $|x|>R$, then the claim holds. However, the OP only assumes that $f\not=0$ for $|x|>R$ (I mixed up the two originally).
Original answer: Yes, this works. If we had $|f(x_N)|\lesssim |x_N|^{-N}$ for arbitrarily large $N$, then the constant coefficient $p=a_0$ in the equation $\sum a_jf^j=0$ solved by $f$ would have the same property.
So it suffices to show that if a polynomial satisfies $p(x)>0$ for $|x|>R$, then in fact $|p(x)|\gtrsim |x|^{-N}$ for some $N$. This follows from the Positivstellensatz, which says, in our setting, that
$$
p(x) = \frac{1 + (|x|^2-R^2)F(x)+G(x)}{(|x|^2-R^2)H(x)+K(x)} \ge \frac{1}{(|x|^2-R^2)H(x)+K(x)} \gtrsim |x|^{-N} ;
$$
here, $F,G, H, K$ are sums of squares of polynomials.
(I suspect there must be a much more elementary argument.)
A: Well it's true in dimension $1$, and I think the following argument
should reduce the case $n>1$ to the $n=1$ statement.
Let $r_0 = \max_{f(x) = 0} \| f(x) \|$.  Then $f$ has constant sign on
$\| x \| > r_0$ (continuous function on a connected set),
and we may assume the sign is positive (else replace $f$ by $-f$).
For $r \geq r_0$, define
$$
F(r) := \min_{\|x\| = r} f(x).
$$
This is an algebraic function of $r$ because it is the value of 
the algebraic function $f(x_1,\ldots,x_n)$ on the algebraic variety 
(typically a curve) in ${\bf R}^{n+2}$ specified by the following
equations on $r$, $x_1,\ldots,x_n$, and $c$:
$$
\sum_{i=1}^n x_i^2 = r^2, \quad
\forall i: \frac{\partial f}{\partial x_i} = c x_i.
$$
But by hypothesis $F(r) \neq 0$ for $r > r_0$, so there exists $N<\infty$
such that $f(r) \gg |r|^{-N}$ for large r, as claimed.
A: Your function $f$ is, in particular, a continuous semi-algebraic function on $\mathbf R^n$, i.e., its graph is a semi-algebraic subset of $\mathbf R^{n+1}$. Such functions are known to have sub-polynomial growth. More precisely (cf. Bochnak, Coste, Roy: Real algebraic geometry, Proposition 2.6.2, p. 43):
Let $S\subseteq\mathbf R^n$ be a closed semi-algebraic subset, and let $f\colon S\rightarrow \mathbf R$ be a continuous semi-algebraic function. Then there are $c\in\mathbf R$ and $N\in\mathbf N$ such that
$$
|f(x)|\leq c(1+||x||^2)^N
$$
for all $x\in S$.
Apply this to the inverse $1/f$ of your function on the closed semi-algebraic subset $||x||\geq A$, where $A\in\mathbf R$ is such that all zeros of $f$ have norm $<A$.
