China Mathematical Olympiad

Proof First, we use induction to prove that $0<a_n\le \frac{1}{2}$, $n=1,2,\dots$. When $n=1$, it is obvious. Now suppose it is true for $n$ ($n\ge 1$), i.e. $0<a_n\le \frac{1}{2}$. Let $f(x)=-x+\frac{1}{2-x}$ for $x\in [0,\frac{1}{2}]$. Then $f(x)$ is a decreasing function. So $$a_{n+1}=f(a_n)\le f(0)=\frac{1}{2},$$ $$a_{n+1}=f(a_n)\ge f(\frac{1}{2})=\frac{1}{6}>0,$$ i.e. it is true for $n+1$.

Back to the problem, it is sufficient to prove: $$\begin{gathered} {\left(\frac{n}{a_1+a_2+\cdots +a_n}\right)}^n{\left(\frac{n}{2(a_1+a_2+\cdots +a_n)}-1\right)}^n \\ \le \left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots \left(\frac{1}{a_n}-1\right). \end{gathered}$$

Let $f(x)=\ln (\frac{1}{x}-1)$ for $x\in (0,\frac{1}{2})$. Then $f(x)$ is a concave function, i.e. for every $0<x_1,x_2<\frac{1}{2}$, we have $f[\frac{x_1+x_2}{2}]\le \frac{f(x_1)+f(x_2)}{2}$. In fact, $$f[\frac{x_1+x_2}{2}]\le \frac{f(x_1)+f(x_2)}{2}$$ is equivalent to $$\begin{gathered} {\left(\frac{2}{x_1+x_2}-1\right)}^2\le \left(\frac{1}{x_1}-1\right)\left(\frac{1}{x_2}-1\right), \\ \iff (x_1-x_2{)}^2\ge 0. \end{gathered}$$ So $f(x)$ is a concave function.

Using Jensen's Inequality, it follows that $$f\left[\frac{x_1+x_2+\cdots +x_n}{n}\right]\le \frac{f(x_1)+f(x_2)+\cdots +f(x_n)}{n},$$ $$\text{so}{\left(\frac{n}{a_1+a_2+\cdots +a_n}-1\right)}^n\le \left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\dots \left(\frac{1}{a_n}-1\right).$$

On the other hand, by using Cauchy's Inequality, we have $$\begin{array}{rl}\sum _{i=1}^n(1-a_i)& =\sum _{i=1}^n\frac{1}{a_i+a_{i+1}}-n\\ & \ge \frac{n^2}{{\sum }_{i=1}^n(a_i+a_{i+1})}-n\\ & =\frac{n^2}{a_{n+1}-a_1+2{\sum }_{i=1}^n{a}_i}-n\\ & \ge \frac{n^2}{2{\sum }_{i=1}^n{a}_i}-n=n\left[\frac{n}{2{\sum }_{i=1}^n{a}_i}-1\right].\end{array}$$ This means $$\frac{{\sum }_{i=1}^n(1-a_i)}{{\sum }_{i=1}^n{a}_i}\ge \frac{n}{{\sum }_{i=1}^n{a}_i}\left[\frac{n}{2{\sum }_{i=1}^n{a}_i}-1\right].$$ So $$\begin{array}{rl}& {\left(\frac{n}{a_1+a_2+\cdots +a_n}\right)}^n{\left(\frac{n}{2(a_1+a_2+\cdots +a_n)}-1\right)}^n\\ & \le {\left[\frac{(1-a_1)(1-a_2)\cdots (1-a_n)}{a_1+a_2+\cdots +a_n}\right]}^n\\ & \le \left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots \left(\frac{1}{a_n}-1\right).\end{array}$$