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.



Every positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?
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.


