, 2023

We are given that $P(x) = x^n Q(1/x)$ and $P(x) \ge Q(x)$ for all real $x$.

First, note that $P(x) = x^n Q(1/x)$ for all $x \ne 0$ (since $Q$ is a polynomial, $Q(1/x)$ is defined for $x \ne 0$).

Also, $P(x) \ge Q(x)$ for all real $x$.

Let us substitute $x$ by $1/x$ (for $x \ne 0$):

$P(1/x) = (1/x)^n Q(x) = x^{-n} Q(x)$.

But $P(1/x) \ge Q(1/x)$ for all $x \ne 0$.

Now, multiply both sides by $x^n$ (for $x > 0$):

$x^n P(1/x) \ge x^n Q(1/x)$.

But $x^n P(1/x) = Q(x)$, and $x^n Q(1/x) = P(x)$, so:

$Q(x) \ge P(x)$ for all $x > 0$.

But from the original condition, $P(x) \ge Q(x)$ for all $x$.

Therefore, for all $x > 0$, $P(x) = Q(x)$.

Since $P$ and $Q$ are polynomials, and two polynomials that agree on an infinite set (here, all $x > 0$) must be equal everywhere, we conclude that $P(x) = Q(x)$ for all real $x$.