Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ? This question was originally asked in stackoverflow (https://math.stackexchange.com/questions/103941/every-positive-polynomial-is-above-a-completely-q-factorized-positive-polynomial) but as it has remained without further feedback for a week  I migrate it here.
Let $P$ be a unitary polynomial with rational coefficients in one variable $x$, such that $P(x) \gt 0$ for all $x \in \mathbb R$. Then $P$ is of even degree, say $2d$. Is it true that there always exist $d$ rational numbers $q_1,q_2, \ldots ,q_d$ such that
$$
(*) \ \ \ P(x) \geq \bigg( \prod_{k=1}^{d} (x-q_k)^2\bigg)
$$
for all $x\in \mathbb R$ ? 
I can show that the answer is YES when $d=1$ or $d=2$.
When $d=1$, P has a canonical form $(X-a)^2+b$ with $b>0$, so we may take
$q_1=a$ ($a$ will be rational since the coefficients of $P$ are) and we are done.
Now assume $d=2$. Then $P$ has a global minimum $\mu_1>0$, attained at one (or several) value $\eta_1$. Then $Q=\frac{P-\mu_1}{(X-\eta_1)^2}$ is a unitary polynomial of degree $2$ in $X$ and is nonnegative everywhere, so we can write
$Q=\mu_2  + (X-\eta_2)^2$ with $\mu_2 \geq 0$. 
   If we write $P$ explicitly as $P=X^4+a_3X^3+a_2X^2+a_1X+a_0$, where 
$a_0,a_1,a_2$ and $a_3$ are rational, then we have
$$
Q=X^2 + (a_3 + 2\eta_1)X + (a_2 + (2a_3\eta_1 + 3\eta_1^2))
$$
So that $\eta_2=-\frac{a_3 + 2\eta_1}{2}$. We deduce the identities
$$
Q=\mu_2+(X+\frac{a_3 + 2\eta_1}{2})^2
$$
$$
P=\mu_1+(X-\eta_1)^2Q=\mu_1+(X-\eta_1)^2\bigg( \mu_2+(X+\frac{a_3 + 2\eta_1}{2})^2\bigg)
$$
Now 
$$
\Omega=\Bigg\lbrace r \in {\mathbb R} \Bigg|  \forall x\in {\mathbb R}, \ P(x) \gt \frac{\mu_1}{2}+(x-r)^2\bigg( \mu_2+(x+\frac{a_3 + 2r}{2})^2\bigg) \Bigg\rbrace
$$
is an open set in $\mathbb R$. It is nonempty, since by construction it contains $\eta_1$. So it will always contain a rational number $q$. Then, we may take $q_1=q$ and $q_2=-\frac{a_3 + 2q}{2}$ and (*) holds.
 A: The function has no real roots, so all its roots are complex numbers, in conjugate pairs. Thus we can factor it into terms of the form $((x-a_i)^2+b_i)$. First consider the product of all the $(x-a_i)^2$. This will be strictly smaller than the original polynomial. The difference will have degree $2d-2$. Our challenge is to replace each $a_i$ with a nearby rational number $d_i$ without making the difference negative anywhere. Since we can choose $d_i$ arbitrarily close to $a_i$, we can make the difference arbitrarily small, so our only concern is its rate of growth. This will be determined by the coefficient of $x^{2d-1}$, which will be $\sum 2(a_i-d_i)$. Since $\sum a_i$ is a rational, we can choose very nearby rationals satisfying $\sum d_i=\sum a_i$, and these will satisfy your inequality.
More formally, the coefficients of $P-\prod (x-d_i)^2$ are continuous functions of $d_i$, and where the $x^{2d-1}$ coefficient is $0$, the minimum value of $(P-\prod(x-d_i)^2)/(1+x^{2d-2})$ is a continuous function of the coefficients. (Adding or subtracting $\epsilon x^k$ changes the result by no more than $\epsilon \max |x^k/(1+x^{2d-2})|.$) So there must be some open ball in the hyperplane where $\sum  a_i=\sum d_i$ where it is still positive. Choose a rational point in that open ball.
A: This question looks as if the classical "catalecticant" (see http://en.wikipedia.org/wiki/Catalecticant) would be relevant.  It is an invariant of binary forms of degree 2n which vanishes if and only if the form is a sum of only n powers.  (I think n=1 will always do but cannot remember.)  For example, a quadratic is a square iff its discriminant is 0;  a quartic is a sum of 2 squares iff its catalecticant (one of the two invariants classically denoted I and J) vanishes.
I seem to remember that Cassels wrote a paper on this, but cannot find it.
A: The answer to your question is yes by the following lemma:
Let $f$ be a polynomial with rational coefficients which is (strictly) positive on the real line and has degree at least $4$. Then there is a nonnegative quadratic polynomial $p$ with rational coefficients such that $f-p$ is nonnegative on the real line and has a multiple rational root.
In my Diplomarbeit (written in German)
http://www.math.uni-konstanz.de/~schweigh/publications/diploma.thesis.pdf
from 1999, I proved this lemma. More precisely I proved ("Satz 2.27" in the Diplomarbeit):
Let $f\in\mathbb R[X]$ be a polynomial of degree $>0$ such that $f(x)>0$ for all $x\in\mathbb R$. Denote by $a$ the smallest global minimizer of $f$. Then there is $\varepsilon>0\in\mathbb R$ such that for all $t\in\mathbb R$ satisfying
$a-\varepsilon< t < a$
$$p_t := f(t)+f'(t)(X-t)+\frac{(f'(t))^2}{4f(t)}(X-t)^2\in\mathbb R[X]$$
is a polynomial of degree 2 such that $p_t \leq f$ on $\mathbb R$.
Note that $p_t$ is a parabola with vanishing discriminant and that $f-p_t$ has a multiple root at $t$ with even multiplicity. If you choose $t$ rational, then $p_t$ has rational coefficients.
In fact, in my Diplomarbeit you find a code which implements in the computer algebra system REDUCE (version 3.6) to find such a $t$. Thus you can compute the $q_k$ in your question using Sturm sequences.
