Team Selection Test for EGMO 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 $x$.