求最大的实数 $C$．使得对任意正整数 $n$ 和满足 $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=1$ 的数列 $\{x_k\}$，均有 $\displaystyle \sum\limits_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right)>C$．
【难度】
【出处】
2017第16届CGMO试题
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【答案】
略
【解析】
首先证明 $C\geqslant\dfrac{1}{3}$．由 $x_{k} \geqslant x_{k-1} \geqslant 0$，得
$\displaystyle \begin{aligned} 3 \sum\limits_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right) &=\sum_{k=1}^{n}\left(x_{k}^{2}+x_{k}^{2}+x_{k}^{2}\right)\left(x_{k}-x_{k-1}\right) >\sum_{k=1}^{n}\left(x_{k}^{2}+x_{k} x_{k-1}+x_{k-1}^{2}\right)\left(x_{k}-x_{k-1}\right) =\sum_{k=1}^{n}\left(x_{k}^{3}-x_{k-1}^{3}\right)=x_{n}^{3}-x_{0}^{3}=1-0=1 \end{aligned}$
下面证明若 $C>\frac{1}{3}$，则存在 $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=1$ 使得 $\displaystyle \sum\limits_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right)<C$．
设 $x_{k}=\frac{k}{n}, k=0,1, \cdots, n$．当 $n>3$ 并且 $\frac{1}{n}<C-\frac{1}{3}$ 时，$\begin{aligned} \sum_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right) &=\dfrac{1}{n^{3}} \sum_{k=1}^{n} k^{2}=\dfrac{n(n+1)(2 n+1)}{6 n^{3}} =\dfrac{1}{3}+\frac{1}{2 n}\left(1+\dfrac{3}{n}\right) <\frac{1}{3}+\dfrac{1}{n}<C \end{aligned}$
因此，$C=\dfrac{1}{3}$．
$\displaystyle \begin{aligned} 3 \sum\limits_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right) &=\sum_{k=1}^{n}\left(x_{k}^{2}+x_{k}^{2}+x_{k}^{2}\right)\left(x_{k}-x_{k-1}\right) >\sum_{k=1}^{n}\left(x_{k}^{2}+x_{k} x_{k-1}+x_{k-1}^{2}\right)\left(x_{k}-x_{k-1}\right) =\sum_{k=1}^{n}\left(x_{k}^{3}-x_{k-1}^{3}\right)=x_{n}^{3}-x_{0}^{3}=1-0=1 \end{aligned}$
下面证明若 $C>\frac{1}{3}$，则存在 $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=1$ 使得 $\displaystyle \sum\limits_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right)<C$．
设 $x_{k}=\frac{k}{n}, k=0,1, \cdots, n$．当 $n>3$ 并且 $\frac{1}{n}<C-\frac{1}{3}$ 时，$\begin{aligned} \sum_{k=1}^{n} x_{k}^{2}\left(x_{k}-x_{k-1}\right) &=\dfrac{1}{n^{3}} \sum_{k=1}^{n} k^{2}=\dfrac{n(n+1)(2 n+1)}{6 n^{3}} =\dfrac{1}{3}+\frac{1}{2 n}\left(1+\dfrac{3}{n}\right) <\frac{1}{3}+\dfrac{1}{n}<C \end{aligned}$
因此，$C=\dfrac{1}{3}$．
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