https://dgrozev.wordpress.com/2023/04/09/bulgarian-nmo-2023-problem-5/

Solved with $\textit{carvilesp7}$ The answer is $2^{1-n}$, achieved when $x_1=2^{1-n}$ and $x_i=2^{i-n-1}$ for every other $i$. It's clear this is true for $n=1$. We will induct on $n$. Let's solve the following problem: we want to find the minimum of \[\frac{|x_{1}|+|x_{1}-x_{2}|+|x_{1}+x_{2}-x_{3}|+\dots +|x_{1}+x_{2}+\dots +x_{n-1}-x_{n}|}{|x_1|+|x_2|+\dots+|x_n|}\]subject to $|x_1|+|x_2|+\dots+|x_n|\not =0$. This is equivalent to the original statement since we can scale everyone by $\frac 1{|x_1|+|x_2|+\dots+|x_n|}$. We will fix $x_1,x_2,\dots,x_{n-1}$ and let $|x_{1}|+|x_{1}-x_{2}|+|x_{1}+x_{2}-x_{3}|+\dots +|x_{1}+x_{2}+\dots +x_{n-2}-x_{n-1}|=p$, $x_1+x_2+\dots+x_{n-1}=s$. We are now trying to find the minimum of \[\frac{p+|s-x_n|}{|x_1|+|x_2|+\dots+|x_n|}\] Notice we can multiply everything by $\pm 1$ to ensure $s\ge0$, also notice if $x_n$ is negative, the value is smaller than if we multiplied it by $-1$ since that moves it closer to $s$, so from now on $x_n\ge 0$. Let $|x_1|+|x_2|+\dots+|x_{n-1}|=a$. Now we want to minimize \[\frac{p+|s-x_n|}{a+x_n}\] $\textbf{Case 1: } x_n\le s$. Our function is equal to $\frac{p+s+a}{a+x_n}-1$, which is decreasing in $x_n$ since $p+s+a>0$. From now on it is enough to consider $x_n\ge s$. $\textbf{Case 2: } x_n\ge s$. Our function is equal to $\frac{p-s-a}{a+x_n}+1$, which is at least 1 if $p>a+s$, which is definitely not the minimum since our conjectured minimum of $2^{1-n}$ is achievable. Otherwise, the function is increasing in $x_n$ meaning it's enough to search for the minimum in the $x_n=s$ situation. Notice, $s\le a$ by triangle inequality, and $\frac pa \ge 2^{2-n}$ by our induction hypothesis, which gives us $\frac p{a+s}\ge \frac p{2a}\ge\frac 12\dot 2^{2-n}=2^{1-n}$.

The answer and equivalency condition is gvole wrote: The answer is $2^{1-n}$, achieved when $x_1=2^{1-n}$ and $x_i=2^{i-n-1}$ for every other $i$. Here's another proof. We induct on $n$, the $n=1$ case is obviously correct. We assume that the $n$ case is correct, consider the $n+1$ case. $|x_1|+\dots+|x_1+\dots+x_{n-1}-x_n|+|x_1+\dots+x_{n}-x_{n+1}|$ $\newline \ge \frac{|x_1|+\dots+|x_n|}{2^{n-1}}+\frac{|x_1+\dots+x_{n}-x_{n+1}|}{2^n}$ $\newline \ge \frac{|x_1|+\dots+|x_n|}{2^{n-1}}+\frac{|x_{n+1}|-|x_1+\dots+x_n|}{2^n}$ $\newline \ge \frac{|x_1|+\dots+|x_n|}{2^{n-1}}+\frac{|x_{n+1}|-(|x_1|+\dots+|x_n|)}{2^n}$ $\newline =\frac{|x_1|+\dots+|x_{n+1}|}{2^n}=2^{-n}$.