IMO Team Selection Test 2, June, 2020

If $x = \alpha$ is a root of $P_n$, then $x = -\alpha$ is a root of $P_n$ as well, as all terms of $P_n$ have even degree. The minimal root of $P_n$ therefore cannot be positive. Therefore $a_n \le 0$ for all $n > 2020$.

We have $P_{n+1}(x) = x^2 \cdot P_n(x) + a_{n+1}$. Substitute $x = a_{n+1}$; as that is a root of $P_n$, we have $P_{n+1}(a_{n+1}) = 0 + a_{n+1} \le 0$.

As the maximal degree term in $P_n(x)$ is $x^{2n}$, there exists an $N < 0$ such that $P_n(x) > 0$ for all $x < N$. Taking for example $-N = \max(2, |a_1| + |a_2| + \dots + |a_n|)$, we see for $x < N$ that $x^{2i-2} \le x^{2n-2}$ for all $1 \le i \le n$ and therefore that
$$\begin{aligned} & |a_1x^{2n-2} + a_2x^{2n-4} + \dots + a_{n-1}x^2 + a_n| \\ & \le |a_1x^{2n-2}| + |a_2x^{2n-4}| + \dots + |a_{n-1}x^2| + |a_n| \\ & \le |a_1x^{2n-2}| + |a_2x^{2n-2}| + \dots + |a_{n-1}x^{2n-2}| + |a_n|x^{2n-2} \\ & \le (|a_1| + |a_2| + \dots + |a_n|)x^{2n-2} \\ & \le -N \cdot x^{2n-2} \\ & < x^{2n}, \end{aligned}$$
so $x^{2n} + a_1x^{2n-2} + a_2x^{2n-4} + \dots + a_{n-1}x^2 + a_n > 0$. Hence for $n \ge 2021$ there exists an $N < 0$ with $P_n(x) > 0$ for all $x < N$, whereas $P_n(a_n) \le 0$. Therefore $P_n(x)$ has a root smaller than $a_n$. As $a_{n+1}$ is the minimal root, we have $a_{n+1} \le a_n$. $\square$