A real number sequence $a_1, \cdots ,a_{2021}$ satisfies the below conditions. $$a_1=1, a_2=2, a_{n+2}=\frac{2a_{n+1}^2}{a_n+a_{n+1}} (1\leq n \leq 2019)$$Let the minimum of $a_1, \cdots ,a_{2021}$ be $m$, and the maximum of $a_1, \cdots ,a_{2021}$ be $M$. Let a 2021 degree polynomial $$P(x):=(x-a_1)(x-a_2) \cdots (x-a_{2021})$$$|P(x)|$ is maximum in $[m, M]$ when $x=\alpha$. Show that $1<\alpha <2$.
Problem
Source: 2021 kmo P5
Tags: Sequence, inequalities, algebra, polynomial
14.11.2021 19:54
What is KMO?
14.11.2021 20:01
KPBY0507 wrote: IMOStarter wrote: What is KMO? Its Korean Mathematics Olympiad Oh, I just think the Kazhastan one
01.12.2021 17:32
Any solution??
01.12.2021 18:01
Since $a_{n+2}-a_{n+1}=\frac{a_{n+1}}{a_n+a_{n+1}}(a_{n+1}-a_{n})$, we can easily show that $a_n$ is increasing and $a_{n+1}-a_n$ is decreasing. It means $m=a_1, M=a_{2021}$. Also, we can prove that $a_{i+k}-a_i< a_{j+k}-a_j$ for $i>j$. Set $\beta = \frac{4}{3}$, then $a_3-a_2=a_2-\beta$. For arbitrary $1<k<2021$ and $a_{k}<x<a_{k+1}$, we will prove that $|P(x)|< |P(\beta)|$. $|P(x)|=(x-a_1)\cdots(x-a_k)(a_{k+1}-x)\cdots(a_{2021}-x)$ For $1<i\leq k$, $x-a_i< a_{k+1}-a_i\leq a_{k-i+3}-a_2\leq a_{k-i+2}-\beta$. For $k+1<i\leq 2021$, $a_i-x< a_i-a_k \leq a_i-\beta$. $\therefore |P(x)|\leq (x-a_1)(a_{k+1}-x)(a_2-\beta)\cdots(a_k-\beta)(a_{k+2}-\beta)\cdots(a_{2021}-\beta)$ Since $\frac{a_1+a_{k+1}}{2}<a_k$, $(x-a_1)(a_{k+1}-x)\leq (a_k-a_1)(a_{k+1}-a_k)$. It is enough to show that $(a_k-a_1)(a_{k+1}-a_k)\leq (\beta-a_1)(a_{k+1}-\beta)=\frac{1}{3}(a_{k+1}-a_1)+\frac{1}{9}$. If $k\geq 4$, $a_{k+1}-a_k \leq \frac{1}{3}$ holds, 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 $k=2, 3$. Hence, We show that $|P(\beta)|>|P(x)|$ for all $a_2<x<M$, so $\alpha$ should exist in the interval $(1, 2)$.
01.12.2021 18:50
Long time no see @KPBY0507