Sorry for the unrelated post, but @oVlad, could you post the other problems or give sources if they are not original (unless they are from shortlists)?

a_507_bc wrote: Sorry for the unrelated post, but @oVlad, could you post the other problems or give sources if they are not original (unless they are from shortlists)? The other problems are (sadly) not new. I'm not even sure whether the ones I posted are original, but I didn't find them anywhere, so I suppose so. Anyway, you can find them at 2023 EGMO TST - Romania.

First observe that if $S=0$, the $LHS$ is equal to $n$. So suppose $S\neq 0$. We can then assume WLOG that $S=1$. We will proceed via smoothing. Let $f(x)=\left|{\frac{1-x}{x}}\right|$. Consider the three intervals: $I=\left(-\infty, 0\right), J=(0,1],K=[1,\infty)$ Suppose none of the $a_i$ are in $I$. Then they are all in $J$, and $\sum_{i=1}^n\left|\frac{1-a_i}{a_i}\right|=\sum_{i=1}^n\frac{1}{a_i}-1\ge n^2-n>\frac{n-1}{n-2}$ from AM-HM (for $n\ge2$). Now we smooth as follows: Let $a$ be $\text{min}(a_1,a_2,\dots,a_n)$. Clearly $a\in I$. Now for any $a_i$ other than $a$ in $I\cup J$, we replace $(a,a_i)\rightarrow (a+a_i-1,1)$. It's clear from looking at the graph that this decreases the LHS. So now we have some $a<0$ and $a_{k_1},a_{k_2},\dots>1$. Note now that $-a\ge a_{k_m}$ for any $m$, and that $f(a)+f(a_{k_m})\ge f(a+\varepsilon)+f(a_{k_m}-\varepsilon)$, so we do this until $(a,a_{k_m})\rightarrow (a+a_{k_m}-1,1)$ and we repeat for all $a_{k_m}$. Then we reach $\left(-(n-2),1,1,\dots,1\right)$, which is the equality case.

oVlad wrote: Let $n\geqslant 3$ be an integer and $a_1,\ldots,a_n$ be nonzero real numbers, with sum $S{}$. Prove that \[\sum_{i=1}^n\left|\frac{S-a_i}{a_i}\right|\geqslant\frac{n-1}{n-2}.\] Are you the author?