Huh, interesting. Clearly $P$ is not constant, so $P(x)\to \pm \infty$ as $x\to\infty$. Assume the former, that is there exists an $M>0$ such that $P(x)>0$ for all $x>M$. Using $P(x)=x^n Q(1/x)$, we obtain $Q(t)>0$ for all $t<1/M$. Now, using $P(t)\ge Q(t)$, we get that $P(t)>0$ for $0<t<1/M$ also. Lastly, $P(1/x) = (1/x)^n Q(x)$ yields, for $x>M$, that $Q(x)>0$ for every $x>M$. Observe now that \[ P(x) P(1/x) = Q(x)Q(1/x). \]So for $x>M$, we know all quantities above are positive, and as $P(x)\ge Q(x)$ and $P(1/x)\ge Q(1/x)$, we must in fact have equality. Since this hols true for infinitely many $x$, we conclude $P(x)=Q(x)$ directly. The case $P(x)\to -\infty$ as $x\to\infty$ should be analogous by sign reversal, although I haven't verified the details. Remark. See below for a much cleaner one-liner: $P(x) = x^n Q(1/x)\ge Q(x) = P(1/x)x^n \ge Q(1/x)x^n$.

$x>0, x^nQ(1/x) \ge Q(x) \ge x^n Q(1/x)$ so they must be equal.