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.
