Pointwise order on polynomials I'm curious about the following problem. Given two polynomials $p,q:\mathbb{R}\rightarrow \mathbb{R}$ is it possible to determine automatically if $p(x)\leq q(x)$ for all $x\in\mathbb{R}$?
I assume, of course, that we are given the expressions for $p$ and $q$:
$p(x)=a_n x^n + \dots a_1 x + a_0$ and $q(x)=b_m x^m + \dots b_1 x + b_0$
with $m,n\in \mathbb{N}$.
If this is possible, is it a consequence of some general theorem that, perhaps, extends to other kinds of (commutative) abstract algebras such as, e.g., the algebra of continuous maps from $X$ compact Hausdorf to $\mathbb{R}$?
Thank you! 
PS: I guess the question boils down to: is it possible to verify if $q-p\geq 0$. Sorry if the question is trivial but I don't know if this is possible. 
 A: As you mentioned, it is enough to consider the problem of when a polynomial $f(x) = c_n x^n + \cdots + c_1 x + c_0$ is always strictly positive.  First of all, you must have $c_n > 0$ (otherwise $f(x) < 0$ for sufficiently large $x$).  Then, by the Intermediate Value Theorem, it is enough to show that $f(x)$ has no real roots.  Any polynomial in $\mathbb{R}[x]$ factors as a product of linear factors and irreducible (over $\mathbb{R}$) quadratics.  Any linear factor corresponds to a root.  So you'll have $f(x) > 0$ for all $x \in \mathbb{R}$ if and only if $f(x)$ is a product of irreducible quadratic polynomials (in particular, this means that the degree of $f(x)$ is even).  Thus, if you completely factor your polynomial, you can tell if it is always positive by looking at the irreducible factors.
A: A real polynomial in one variable is positive if and only if it is a sum of squares of two real polynomials. A proof is given in Proposition 1.2.1 here. I'm not sure whether that leads to an algorithm for determining whether a given real polynomial in one variable is positive, but it looks like a good start. 
