实数列 $\{a_n\}$ 满足 $a_{1}=\dfrac{1}{2},a_{k+1}=-a_{k}+\dfrac{1}{2-a_{k}}, k=1,2, \cdots$
证明：不等式 $\left(\dfrac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right)^{n} \leqslant\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
证明：不等式 $\left(\dfrac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right)^{n} \leqslant\left(\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
【难度】
【出处】
2006第21届CMO试题
【标注】
【答案】
略
【解析】
首先，用数学归纳法证明 $0<a_{n} \leqslant \frac{1}{2}, n=1,2, \cdots,n=1$ 时，命题显然成立．
假设命题对 $n(n\geqslant 1)$ 成立，即有 $0<a_{n} \leqslant \dfrac{1}{2}$．
设 $f(x)=-x+\frac{1}{2-x}, x \in\left[0, \frac{1}{2}\right]$，则 $f(x)$ 是减函数，于是 $a_{n+1}=f\left(a_{n}\right) \leqslant f(0)=\frac{1}{2},a_{n+1}=f\left(a_{n}\right) \geqslant f\left(\frac{1}{2}\right)=\frac{1}{6}>0$
即命题对 $n + 1$ 也成立原命题等价于 $\left(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n}\left(\dfrac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
设 $f(x)=\ln \left(\dfrac{1}{x}-1\right), x \in\left(0, \dfrac{1}{2}\right)$，则 $f(x)$ 是凹函数，即对 $0<x_{1}, x_{2}<\dfrac{1}{2}$，有 $f\left(\dfrac{x_{1}+x_{2}}{2}\right) \leqslant \dfrac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}$
事实上 $f\left(\dfrac{x_{1}+x_{2}}{2}\right) \leqslant \dfrac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}$
等价于 $\begin{aligned}\left(\dfrac{2}{x_{1}+x_{2}}-1\right)^{2} \leqslant\left(\dfrac{1}{x_{1}}-1\right)\left(\frac{1}{x_{2}}-1\right) & \Leftrightarrow \left(x_{1}-x_{2}\right)^{2} \geqslant 0 \end{aligned}$
所以，由Jenson 不等式可得 $f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) \leqslant \dfrac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{a}\right)}{n}$
即 $\left(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}-1\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
另一方面，由题设及柯西不等式，可得
$\displaystyle \sum\limits_{i=1}^{n}\left(1-a_{i}\right)=\sum_{i=1}^{n} \frac{1}{a_{i}+a_{i+1}}-n \geqslant \frac{n^{2}}{\sum\limits_{i=1}^{n}\left(a_{i}+a_{i+1}\right)}-n=\frac{n^{2}}{a_{n+1}-a_{1}+2 \sum\limits_{i=1}^{n} a_{i}}-n \geqslant\frac{n^{2}}{2 \sum\limits_{i=1}^{n} a_{i}}-n=n\left[\frac{n}{2 \sum\limits_{i=1}^{n} a_{i}}-1\right]$ 所以 $\displaystyle \frac{\sum\limits_{i=1}^{n}\left(1-a_{i}\right)}{\sum\limits_{i=1}^{n} a_{i}} \geqslant \frac{n}{\sum\limits_{i=1}^{n} a_{i}}\left[\frac{n}{2 \sum\limits_{i=1}^{n} a_{i}}-1\right]$
故 $\left(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n}\left(\dfrac{n}{2} \dfrac{n}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right)^{n} \leqslant\left(\dfrac{\left(1-a_{1}\right)+\left(1-a_{2}\right)+\cdots+\left(1-a_{n}\right)}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
从而原命题得证．
假设命题对 $n(n\geqslant 1)$ 成立，即有 $0<a_{n} \leqslant \dfrac{1}{2}$．
设 $f(x)=-x+\frac{1}{2-x}, x \in\left[0, \frac{1}{2}\right]$，则 $f(x)$ 是减函数，于是 $a_{n+1}=f\left(a_{n}\right) \leqslant f(0)=\frac{1}{2},a_{n+1}=f\left(a_{n}\right) \geqslant f\left(\frac{1}{2}\right)=\frac{1}{6}>0$
即命题对 $n + 1$ 也成立原命题等价于 $\left(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n}\left(\dfrac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
设 $f(x)=\ln \left(\dfrac{1}{x}-1\right), x \in\left(0, \dfrac{1}{2}\right)$，则 $f(x)$ 是凹函数，即对 $0<x_{1}, x_{2}<\dfrac{1}{2}$，有 $f\left(\dfrac{x_{1}+x_{2}}{2}\right) \leqslant \dfrac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}$
事实上 $f\left(\dfrac{x_{1}+x_{2}}{2}\right) \leqslant \dfrac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2}$
等价于 $\begin{aligned}\left(\dfrac{2}{x_{1}+x_{2}}-1\right)^{2} \leqslant\left(\dfrac{1}{x_{1}}-1\right)\left(\frac{1}{x_{2}}-1\right) & \Leftrightarrow \left(x_{1}-x_{2}\right)^{2} \geqslant 0 \end{aligned}$
所以，由Jenson 不等式可得 $f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right) \leqslant \dfrac{f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{a}\right)}{n}$
即 $\left(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}-1\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
另一方面，由题设及柯西不等式，可得
$\displaystyle \sum\limits_{i=1}^{n}\left(1-a_{i}\right)=\sum_{i=1}^{n} \frac{1}{a_{i}+a_{i+1}}-n \geqslant \frac{n^{2}}{\sum\limits_{i=1}^{n}\left(a_{i}+a_{i+1}\right)}-n=\frac{n^{2}}{a_{n+1}-a_{1}+2 \sum\limits_{i=1}^{n} a_{i}}-n \geqslant\frac{n^{2}}{2 \sum\limits_{i=1}^{n} a_{i}}-n=n\left[\frac{n}{2 \sum\limits_{i=1}^{n} a_{i}}-1\right]$ 所以 $\displaystyle \frac{\sum\limits_{i=1}^{n}\left(1-a_{i}\right)}{\sum\limits_{i=1}^{n} a_{i}} \geqslant \frac{n}{\sum\limits_{i=1}^{n} a_{i}}\left[\frac{n}{2 \sum\limits_{i=1}^{n} a_{i}}-1\right]$
故 $\left(\dfrac{n}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n}\left(\dfrac{n}{2} \dfrac{n}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right)^{n} \leqslant\left(\dfrac{\left(1-a_{1}\right)+\left(1-a_{2}\right)+\cdots+\left(1-a_{n}\right)}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n} \leqslant\left(\dfrac{1}{a_{1}}-1\right)\left(\dfrac{1}{a_{2}}-1\right) \cdots\left(\dfrac{1}{a_{n}}-1\right)$
从而原命题得证．
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