For a positive integer $n$ and real numbers $a_1, a_2, \dots ,a_n$ we'll define $b_1, b_2, \dots ,b_{n+1}$ such that $b_k=a_k+\max({a_{k+1},a_{k+2}})$ for all $1\leq k \leq n$ and $b_{n+1}=b_1$. (Also $a_{n+1}=a_1$ and $a_{n+2}=a_2$) Find the least possible value of $\lambda$ such that for all $n, a_1, \dots, a_n$ the inequality $$\lambda \Biggl[ \sum_{i=1}^n(a_i-a_{i+1})^{2024} \Biggr] \geq \sum_{i=1}^n(b_i-b_{i+1})^{2024}$$holds.
Problem
Source: 2024 Turkey TST P6
Tags: inequalities
19.03.2024 11:07
Define $d_i=a_{i+1}-a_i$. Then $b_{i+1}-b_i=d_i+\min(d_{i+1},0)+\max(d_{i+2},0)$. Use $M_1\leq M_{2024}$ to obtain $(x+y)^{2024}\leq 2^{2023}(x^{2024}+y^{2024})$. Also, $(u+v)^{2024}\leq u^{2024}+v^{2024}$ for a non-positive $u$ and a non-negative $v$ because $|u-v|\leq |u|$ and $|u-v|\leq |v|$. Use these and $$\sum (d_i+\min(d_{i+1},0)+\max(d_{i+2},0))^{2024}\leq \sum 2^{2023}(d_i^{2024}+(\min(d_{i+1},0)+\max(d_{i+2},0))^{2024})\leq\sum 2^{2023}(d_i^{2024}+\min(d_{i+1},0)^{2024}+\max(d_{i+2},0)^{2024})$$And $$\sum 2^{2023}(d_i^{2024}+\min(d_{i+1},0)^{2024}+\max(d_{i+2},0)^{2024})=\sum 2^{2023}(d_i^{2024}+\min(d_i,0)^{2024}+\max(d_i,0)^{2024})=2^{2024}\sum d_i^{2024}$$That is, $\lambda$ is at most $2^{2024}$. Now, for $n=2m$, take $\{d_i\}=\{1,1,\cdots,1,-1,-1,\cdots,-1\}$. Then, $\lambda$ has to satisfy the condition $$\lambda\cdot n\geq 2^{2024}(n-4)+1+1+0+0$$for every $n$. That means $\lambda$ is at least $2^{2024}$. So the answer is $\lambda=2^{2024}$
19.03.2024 11:30
20.03.2024 06:01
Quite a beautiful problem!
01.05.2024 05:53
I learnt this problem at ns in Shanghai, a beautiful but easy question !