实数 $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ 满足：$a_{1}+a_{2}+\cdots+a_{n}=0$，且 $2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots, n-1$．求最小的 $\lambda(n)$，使得对所有 $k \in\{1,2, \cdots, n\}$，都有 $\left|a_{k}\right| \leqslant \lambda(n) \cdot \max \left\{\left|a_{1}\right|,\left|a_{n}\right|\right\}$．
【难度】
【出处】
2009年中国西部数学奥林匹克试题
【标注】
【答案】
略
【解析】
$\lambda(n)_{\min }=\frac{n \pm 1}{n-1}$
首先，取 $a_{1}=1, a_{2}=-\dfrac{n+1}{n-1},a_{k}=-\dfrac{n+1}{n-1}+\dfrac{2 n(k-2)}{(n-1)(n-2)}, k=3,4, \cdots, n$，则满足 $a_{1}+a_{2}+\cdots+a_{n}=0$ 及 $2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots,n-1$．此时 $\lambda(n) \geqslant \dfrac{n+1}{n-1}$．
下证 $\lambda(n)=\dfrac{n+1}{n-1}$ 时，对所有 $k \in\{1,2, \cdots, n\}$ 都有 $\left|a_{k}\right| \leqslant \lambda(n) \cdot \max \left\{\left|a_{1}\right|,\left|a_{n}\right|\right\}$．
因此 $2 a_{k} \leqslant a_{k-1}+a_{k+1}$，所以 $a_{k+1}-a_{k} \geqslant a_{k}-a_{k-1}$，于是 $a_{n}-a_{n-1} \geqslant a_{n-1}-a_{n-2} \geqslant \cdots \geqslant a_{2}-a_{1}$，所以
$\begin{aligned} &(k-1)\left(a_{n}-a_{1}\right) \\=&(k-1)\left[\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)\right] \\ \geqslant &(n-1)\left[\left(a_{k}-a_{k-1}\right)+\left(a_{k-1}-a_{k-2}\right)+\cdots+\left(a_{2}-a_{1}\right)\right] \\=&(n-1)\left(a_{k}-a_{1}\right) \end{aligned}$
故 $\begin{aligned} a_{k} & \leqslant \frac{k-1}{n-1}\left(a_{n}-a_{1}\right)+a_{1} =\frac{1}{n-1}\left[(k-1) a_{n}+(n-k) a_{1}\right] \end{aligned}$ ①
同 ① 可得，对固定的 $k$，且 $k\ne 1,n$，当 $\leqslant j\leqslant k$ 时，$a_{j} \leqslant \dfrac{1}{k-1}\left[(j-1) a_{k}+(k-j) a_{1}\right]$，当 $k\leqslant j\leqslant n$ 时，$a_{j} \leqslant \dfrac{1}{n-k}\left[(j-k) a_{n}+(n-j) a_{k}\right]$，所以
$\begin{aligned} \sum_{j=1}^{k} a_{j} & \leqslant \frac{1}{k-1} \sum_{j=1}^{k}\left[(j-1) a_{k}+(k-j) a_{1}\right] \\ &=\frac{k}{2}\left(a_{1}+a_{k}\right) \\ \sum_{j=k}^{n} a_{j} & \leqslant \frac{1}{n-k} \sum_{j=k}^{n}\left[(j-k) a_{n}+(n-j) a_{k}\right] \\ &=\frac{n+1-k}{2}\left(a_{k}+a_{n}\right) \end{aligned}$
相加得 $\begin{aligned} a_{k} &=\sum_{j=1}^{k} a_{j}+\sum_{j=k}^{n} a_{j} \leqslant \frac{k}{2}\left(a_{1}+a_{k}\right)+\frac{n+1-k}{2}\left(a_{k}+a_{n}\right)=\frac{k}{2} a_{1}+\frac{n+1}{2} a_{k}+\frac{n+1-k}{2} a_{n} \end{aligned}$，
所以 $a_{k} \geqslant-\dfrac{1}{n-1}\left[k a_{1}+(n+1-k) a_{n}\right]$ ②
由 ①，② 得
$\begin{aligned}\left|a_{k}\right| \leqslant & \max \left\{\dfrac{1}{n-1}\left|(k-1) a_{n}+(n-k) a_{1}\right|, \dfrac{1}{n-1}\left|k a_{1}+(n+1-k) a_{n}\right| \right\} \\ \leqslant & \dfrac{n+1}{n-1} \max \left\{\left|a_{1}\right|,\left|a_{n}\right|\right\}, k=2,3, \cdots, n-1 \end{aligned}$
综上所述，$\lambda(n)_{\min }=\dfrac{n+1}{n-1}$．
首先，取 $a_{1}=1, a_{2}=-\dfrac{n+1}{n-1},a_{k}=-\dfrac{n+1}{n-1}+\dfrac{2 n(k-2)}{(n-1)(n-2)}, k=3,4, \cdots, n$，则满足 $a_{1}+a_{2}+\cdots+a_{n}=0$ 及 $2 a_{k} \leqslant a_{k-1}+a_{k+1}, k=2,3, \cdots,n-1$．此时 $\lambda(n) \geqslant \dfrac{n+1}{n-1}$．
下证 $\lambda(n)=\dfrac{n+1}{n-1}$ 时，对所有 $k \in\{1,2, \cdots, n\}$ 都有 $\left|a_{k}\right| \leqslant \lambda(n) \cdot \max \left\{\left|a_{1}\right|,\left|a_{n}\right|\right\}$．
因此 $2 a_{k} \leqslant a_{k-1}+a_{k+1}$，所以 $a_{k+1}-a_{k} \geqslant a_{k}-a_{k-1}$，于是 $a_{n}-a_{n-1} \geqslant a_{n-1}-a_{n-2} \geqslant \cdots \geqslant a_{2}-a_{1}$，所以
$\begin{aligned} &(k-1)\left(a_{n}-a_{1}\right) \\=&(k-1)\left[\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right)\right] \\ \geqslant &(n-1)\left[\left(a_{k}-a_{k-1}\right)+\left(a_{k-1}-a_{k-2}\right)+\cdots+\left(a_{2}-a_{1}\right)\right] \\=&(n-1)\left(a_{k}-a_{1}\right) \end{aligned}$
故 $\begin{aligned} a_{k} & \leqslant \frac{k-1}{n-1}\left(a_{n}-a_{1}\right)+a_{1} =\frac{1}{n-1}\left[(k-1) a_{n}+(n-k) a_{1}\right] \end{aligned}$ ①
同 ① 可得，对固定的 $k$，且 $k\ne 1,n$，当 $\leqslant j\leqslant k$ 时，$a_{j} \leqslant \dfrac{1}{k-1}\left[(j-1) a_{k}+(k-j) a_{1}\right]$，当 $k\leqslant j\leqslant n$ 时，$a_{j} \leqslant \dfrac{1}{n-k}\left[(j-k) a_{n}+(n-j) a_{k}\right]$，所以
$\begin{aligned} \sum_{j=1}^{k} a_{j} & \leqslant \frac{1}{k-1} \sum_{j=1}^{k}\left[(j-1) a_{k}+(k-j) a_{1}\right] \\ &=\frac{k}{2}\left(a_{1}+a_{k}\right) \\ \sum_{j=k}^{n} a_{j} & \leqslant \frac{1}{n-k} \sum_{j=k}^{n}\left[(j-k) a_{n}+(n-j) a_{k}\right] \\ &=\frac{n+1-k}{2}\left(a_{k}+a_{n}\right) \end{aligned}$
相加得 $\begin{aligned} a_{k} &=\sum_{j=1}^{k} a_{j}+\sum_{j=k}^{n} a_{j} \leqslant \frac{k}{2}\left(a_{1}+a_{k}\right)+\frac{n+1-k}{2}\left(a_{k}+a_{n}\right)=\frac{k}{2} a_{1}+\frac{n+1}{2} a_{k}+\frac{n+1-k}{2} a_{n} \end{aligned}$，
所以 $a_{k} \geqslant-\dfrac{1}{n-1}\left[k a_{1}+(n+1-k) a_{n}\right]$ ②
由 ①，② 得
$\begin{aligned}\left|a_{k}\right| \leqslant & \max \left\{\dfrac{1}{n-1}\left|(k-1) a_{n}+(n-k) a_{1}\right|, \dfrac{1}{n-1}\left|k a_{1}+(n+1-k) a_{n}\right| \right\} \\ \leqslant & \dfrac{n+1}{n-1} \max \left\{\left|a_{1}\right|,\left|a_{n}\right|\right\}, k=2,3, \cdots, n-1 \end{aligned}$
综上所述，$\lambda(n)_{\min }=\dfrac{n+1}{n-1}$．
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