Let $a_1,a_2,\cdots,a_{100}\geq 0$ such that $\max\{a_{i-1}+a_i,a_i+a_{i+1}\}\geq i $ for any $2\leq i\leq 99.$ Find the minimum of $a_1+a_2+\cdots+a_{100}.$
Problem
Source: China Western Mathematical Olympiad 2023 Day 2 P1
Tags: inequalities, algebra, China
07.08.2023 14:52
The answer is $\boxed{1683}$. $$x_i=\begin{cases}i+1\quad i\equiv 2\pmod 3\\0\quad\text{others}\end{cases}$$
07.08.2023 14:55
P.S.Try to solve the minimum of $a_1^2+a_2^2+\cdots +a_{100}^2$.
07.08.2023 15:13
07.08.2023 15:15
Let $a_1,a_2,\cdots,a_{100}\geq 0$ such that $\max\{a_{i-1}+a_i,a_i+a_{i+1}\}\geq i $ for any $2\leq i\leq 99.$ Find the minimum of $a_1a_2+a_2a_3+\cdots+a_{99}a_{100}+a_{100}a_1.$
08.08.2023 03:47
sqing wrote: Let $a_1,a_2,\cdots,a_{100}\geq 0$ such that $\max\{a_{i-1}+a_i,a_i+a_{i+1}\}\geq i $ for any $2\leq i\leq 99.$ Find the minimum of $a_1a_2+a_2a_3+\cdots+a_{99}a_{100}+a_{100}a_1.$ The answer is $\boxed{0}$.Holds the same as above.(By my friend ZJL)
13.05.2024 03:35
PhilippineMonkey wrote: The answer is $\boxed{1683}$. $$x_i=\begin{cases}i+1\quad i\equiv 2\pmod 3\\0\quad\text{others}\end{cases}$$ Why is it the minimum?
13.05.2024 05:17