Problem 1569

1. Given Conditions and Initial Setup:
We are given a sequence $a_1, a_2, \ldots, a_{2021}$ defined by:
$$a_1 = 1, \quad a_2 = 2, \quad a_{n+2} = \frac{2a_{n+1}^2}{a_n + a_{n+1}} \quad \text{for} \quad 1 \leq n \leq 2019$$
We need to show that the maximum value of $|P(x)|$ in the interval $[m, M]$ occurs at some $\alpha$ such that $1 < \alpha < 2$, where $m$ and $M$ are the minimum and maximum values of the sequence, respectively.

2. Monotonicity and Bounds:
We first show that the sequence $a_n$ is increasing and that $a_{n+1} - a_n$ is decreasing. This can be seen from the recurrence relation:
$$a_{n+2} - a_{n+1} = \frac{2a_{n+1}^2}{a_n + a_{n+1}} - a_{n+1} = \frac{a_{n+1}(a_{n+1} - a_n)}{a_n + a_{n+1}}$$
Since $a_{n+1} > a_n$, the term $\frac{a_{n+1}(a_{n+1} - a_n)}{a_n + a_{n+1}}$ is positive, implying $a_{n+2} > a_{n+1}$. Thus, the sequence is increasing.

3. Minimum and Maximum Values:
Since the sequence is increasing, the minimum value $m$ is $a_1 = 1$ and the maximum value $M$ is $a_{2021}$.

4. **Polynomial $P(x)$:**
Consider the polynomial:
$$P(x) = (x - a_1)(x - a_2) \cdots (x - a_{2021})$$
We need to show that $|P(x)|$ is maximized at some $\alpha$ such that $1 < \alpha < 2$.

5. **Intermediate Value $\beta$:**
Set $\beta = \frac{4}{3}$. We will show that $|P(x)| < |P(\beta)|$ for any $a_2 < x < M$.

6. **Bounding $|P(x)|$:**
For $1 < k < 2021$ and $a_k < x < a_{k+1}$, we have:
$$|P(x)| = (x - a_1)(x - a_2) \cdots (x - a_k)(a_{k+1} - x) \cdots (a_{2021} - x)$$
For $1 \leq i \leq k$, $x - a_i < a_{k+1} - a_i \leq a_{k-i+3} - a_2 \leq a_{k-i+2} - \beta$.
For $k+1 \leq i \leq 2021$, $a_i - x < a_i - a_k \leq a_i - \beta$.

7. Combining Inequalities:
Therefore,
$$|P(x)| \leq (x - a_1)(a_{k+1} - x)(a_2 - \beta) \cdots (a_k - \beta)(a_{k+2} - \beta) \cdots (a_{2021} - \beta)$$
Since $\frac{a_1 + a_{k+1}}{2} < a_k$, we have:
$$(x - a_1)(a_{k+1} - x) \leq (a_k - a_1)(a_{k+1} - a_k)$$

8. Final Inequality:
It is sufficient to show that:
$$(a_k - a_1)(a_{k+1} - a_k) \leq (\beta - a_1)(a_{k+1} - \beta) = \frac{1}{3}(a_{k+1} - a_1) + \frac{1}{9}$$
For $k \geq 4$, $a_{k+1} - a_k \leq \frac{1}{3}$, so:
$$(a_k - a_1)(a_{k+1} - a_k) \leq \frac{1}{3}(a_k - a_1) < \frac{1}{3}(a_{k+1} - a_1) + \frac{1}{9}$$
This inequality also holds for $k = 2, 3$.

9. Conclusion:
Hence, we have shown that $|P(\beta)| > |P(x)|$ for all $a_2 < x < M$, so $\alpha$ should exist in the interval $(1, 2)$.