let $a_1,a_2,...a_n$ a sequence of real numbers such that $a_1+....+a_n=0$. define $b_i=a_1+a_2+....a_i$ for all $1 \leq i \leq n$ .suppose $b_i(a_{j+1}-a_{i+1}) \geq 0$ for all $1 \leq i \leq j \leq n-1$. Show that $$\max_{1 \leq l \leq n} |a_l| \geq \max_{1 \leq m \leq n} |b_m|$$
Problem
Source: Netherlands TST for IMO 2017 day 3 problem 2
Tags: Sequences, algebra
Math-Ninja
01.02.2018 18:16
bump......
Medjl
01.02.2018 18:17
Math-Ninja wrote: bump...... why are you bumping the post if you want a solution i can give it to you ?
Math-Ninja
01.02.2018 18:22
Yes please... I thought you wanted help
ThE-dArK-lOrD
01.02.2018 19:39
Let $m$ is the smallest index that maximize $|b_m|$. Note that if $|b_m|=0$, we get that all terms $a_1,a_2,...,a_n$ must be zero and the inequality clearly holds. Note that multiplying every terms of $a_1,a_2,...,a_n$ by $-1$ doesn't change the problem. Hence, we can WLOG assume that $b_m\geq 0$. Since $b_m\geq |b_{m+1}|$ and $b_m\geq |b_{m-1}|$, we get that $a_{m+1}\leq 0$ and $a_{m}\geq 0$. But if $a_m=0$, we get that $b_m=b_{m-1}$, contradiction to the definition of $m$. So, $a_m>0$. We get that $a_m>a_{m+1}\implies a_{m+1}-a_m<0$. Since $b_{m-1}(a_{m+1}-a_m)\geq 0$, we get that $b_{m-1}\leq 0$. Hence, $a_m=b_m-b_{m-1}\geq b_m$. This gives $\max_{1\leq \ell \leq n}{|a_{\ell} |} \geq |a_m| =a_m\geq b_m$, done.