2 corrected spelling in title

1

# EEvery positive polynomial with rational coefficients is above a completely Q-factorized nonnegative polynomial ?

This question was originally asked in stackoverflow (http://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.