horizon wrote: The real numbers $a_{1},a_{2},\ldots ,a_{n}$ where $n\ge 3$ are such that $\sum_{i=1}^{n}a_{i}=0$ and $2a_{k}\le\ a_{k-1}+a_{k+1}$ for all $k=2,3,\ldots ,n-1$. Find the least $f(n)$ such that, for all $k\in\left\{1,2,\ldots ,n\right\}$, we have $|a_{k}|\le f(n)\max\left\{|a_{1}|,|a_{n}|\right\}$ So $2,0,-1,-2,-1,0,2$ for $a_1$ to $a_7$ satisfy and $0\ge 2 f(7)$ and $f(7)=0$ will always work. (similar for other $f(n)$ seems to be a stupid solution for a Chinese P8)

$\frac{n+1}{n-1}$

My solution First take $a_1=1 $,$ a_2=\frac{n+1}{n-1} $ and $a_k=-\frac{n+1}{n-1}+\frac{2n(k-2)}{(n-1)(n-2)}$ we can get $f(n)\ge \frac{n+1}{n-1}$ Now I claim that the minimun of $f(n)$ is $\frac{n+1}{n-1}$ Proof Notice that sequence $a_i$ is convex so we have $a_k\le{\frac{1}{n-1}[(k-1)a_n+(n-k)a_1]}……(1)$ and for a fixed k (not equal to $1$ and $n$) we have $a_j\le{\frac{1}{k-1}[(j-1)a_k+(k-j)a_1]}$ when $1\le{j}\le{k}$ and $a_j\le{\frac{1}{n-k}[(j-k)a_n+(n-j)a_k]}$ when $k\le{j}\le{n}$ hence we have $\sum_{j=1}^{k}a_j\le{\frac{k}{2}(a_1+a_k)}$ and $\sum_{j=k}^{n}\le{\frac{n+1-k}{2}(a_k+a_n)}$ so we obtain $a_k\le{\frac{k}{2}a_1+\frac{n+1}{2}a_k+\frac{n+1-k}{2}a_n}$ then $a_k\ge {-\frac{1}{n-1}[ka_1+(n+1-k)a_n)]}$$……$$ (2)$ from $(1)$and $ (2)$we get the result .We are done

The above solution is very similar to the official one; they just differ from the “convex sequence” part. Anyways, this is a weird problem and I don’t think their will be many methods.