Lemma Let$0<x_1,x_2,..x_n<\frac{1}{2}$we have$(\frac{1}{x_1}-1)(\frac{1}{x_2}-1)..(\frac{1}{x_n}-1)\geq \left(\frac{n}{x_1+x_2+..+x_n}-1\right)^n$. Proof by induction. Put$\alpha=1-\sqrt\frac{1}{2}\rightarrow 2\alpha(2-\alpha)=1$It's easy to prove that$a_{2k+1}>\alpha>a_{2k};a_{2k+1}+a_{2k}>2\alpha\forall k$The remain is obvious

At least for me the rest wasn't so obvious as mumath said.So i will post a complete solution: We will prove by induction that $ a_i > a_{i + 1}$.Indeed,$ a_i > a_{i + 1}\Longleftrightarrow 2a_{i} > \frac {1}{2 - a_i}$,but by induction principle: $ a_i < a_{i - 1} < \dots < a_1 = \frac {1}{2}$,so our inequality is proved,as a consequence we've just proved that $ a_i\leq\frac {1}{2}$. Also $ - a_k + \frac {1}{2 - a_k} = \frac {a_k^2 - 2a_{k} + 1}{2 - a_k} = \frac {(a_k - 1)^2}{2 - a_k}$,hence $ a_k > 0$,so finally: $ 0 < a_k\leq\frac {1}{2}$,for all $ k\geq 1$. As mumath showed we can apply Jensen inequality to obtain: $ (\frac {1}{a_1} - 1)\dots(\frac {1}{a_n} - 1)\geq(\frac {n}{a_1 + a_2 + \dots + a_n} - 1)^n$. So we need to prove that $ \boxed{(\frac {n}{a_1 + a_2 + \dots + a_n} - 1)\geq(\frac {n}{2(a_1 + a_2\dots + a_n)} - 1)(\frac {n}{a_1 + a_2 + \dots + a_n})}$ Let $ A = \sum_{i = 1}^{n}a_i$.It is easy to understand that: $ 2A = (a_1 + a_2) + (a_2 + a_3) + \dots + (a_{n - 1} + a_n) + (a_n + a_{n + 1}) + a_1 - a_{n + 1} = \frac {1}{2 - a_1} + \frac {1}{2 - a_2} + \dots + \frac {1}{2 - a_{n}} + a_1 - a_{n + 1}$. Using CBS inequality: $ \frac {1}{2 - a_1} + \frac {1}{2 - a_2} + \dots + \frac {1}{2 - a_{n}} + a_1 - a_{n + 1}\geq\frac {n^2}{2n - A} + a_1 - a_{n + 1}$ But as we've proved that $ a_i$ is decreasing sequence,it follows that $ a_{n + 1} < a_1$,so $ \frac {n^2}{2n - A} + a_1 - a_{n + 1}\geq \frac {n^2}{2n - A}$. Hence we conclude that $ 2A(2n - A)\geq n^2$. Our boxed inequality is equivalent to the following: $ \frac {n - A}{A}\geq(\frac {n}{A})(\frac {n - 2A}{2A})$,which is equivalent to: $ 2A(2n - A)\geq n^2$.But we've already proved this inequality,so we are done.

Hello everybody I'm new in this forum, and I wanted to post another solution to this problem So first, let's get show that $a_k<1$, for $k=1,2,...$. For $k=1$, we have $a_1=\frac{1}{2}<1.$Then, if we suppose that $a_n<1$, we get :$$a_n(a_n-1)<1\Leftrightarrow (1-a_n)(2-a_n)<1\Leftrightarrow \frac{1}{2-a_n}<1+a_n\Leftrightarrow a_{n+1}<1$$ and we close the induction. Then, we get : $$\sum_{k=1}^n \frac{a_k}{1-a_k}=\sum_{k=1}^n \frac{1}{\frac{1}{a_k}-1}\ge 0. $$ Then we use GM-HM inequality and we get : $\sqrt[n\,]{\displaystyle\prod_{k=1}^n \frac{1}{a_k}-1}\ge\frac{n}{\displaystyle\sum_{k=1}^n \frac{1}{\frac{1}{a_k}-1}}\ge 0 $ So $\frac{2}{n^2}\bigg(\sum_{k=1}^n a_k \bigg)^2\sqrt[n\,]{\displaystyle\prod_{k=1}^n \frac{1}{a_k}-1}\ge 0$ Therefore $\frac{n}{2}\cdot\frac{1}{\displaystyle\sum_{k=1}^n a_k}-1\le\frac{n}{2}\cdot\frac{1}{\displaystyle\sum_{k=1}^n a_k}\le\frac{1}{n}\displaystyle\sum_{k=1}^n \sqrt[n\,]{\displaystyle\prod_{k=1}^n \frac{1}{a_k}-1} $ And finally we have $\left(\frac{n}{2(a_1+a_2+\cdots+a_n)}-1\right)^n \leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots \left(\frac{1}{a_n}-1\right).$

Erken wrote: At least for me the rest wasn't so obvious as mumath said. So i will post a complete solution: We will prove by induction that $a_i>a_{i-1}$. Indeed, $a_i>a_{i-1}\Longleftrightarrow 2a_i>\frac{1}{2-a_i}$, but by induction principle: $a_i<a_{i-1}<\dots<a_1=\frac{1}{2}$, so our inequality is proved, as a consequence we've just proved that $a_i\leq\frac{1}{2}$. It seems Erken didn't kind of really get mumath's point, and this beginning part of proof is also incorrect. It feels a bit disturbing to just leave the posts like this, thus let me add a correction as well as some remarks. First note that $a<\frac{1}{2}$ doesn't guarantee $2a>\frac{1}{2-a}$ and $a_i$ doesn't monotonically decrease... Instead we actually have \[a_2<a_4<...<a_{2i}<1-\frac{1}{\sqrt{2}}<a_{2i-1}<...<a_3<a_1\]and $\frac{a_{2i}+a_{2i+1}}{2}<1-\frac{1}{\sqrt{2}}<\frac{a_{2i-1}+a_{2i}}{2}$ for each $i\geq 1$. The reason is, $f=-x+\frac{1}{2-x}$ is monotonically decreasing over $[0,1]$ and a contraction mapping (easy to verify that it has Lipschitz constant $3/4$). This furthermore guarantees that the fixed point $1-\frac{1}{\sqrt{2}}\in [0,1]$ of $f$ exists and is unique, by the famous Banach fixed-point theorem. Consequently we have \[1-\frac{1}{\sqrt{2}}<\frac{a_1+...a_n}{n}\leq a_1\]for each $n\geq 1$, which implies $\left(\frac {n}{2(a_1+a_2+...+a_n)}-1\right)<\left(\frac{a_1+a_2+...+a_n}{n}\right)\left(\frac {n}{a_1+a_2+...+a_n}-1\right)$. The point is, given any decreasing function $f$ from some closed interval to itself, once you know that $f$ is a contraction mapping and has $\alpha$ as the fixed point, then any sequence defined by $a_{k+1}=f(a_k)$ starting from some $a_1>\alpha$ in the interval will satisfy the inequalities above with $1-\frac{1}{\sqrt{2}}$ replaced by $\alpha$ and the last one replaced by $\frac{a_1+...a_n}{n}<f(\frac{a_1+...a_n}{n})$. I believe the author Prof. Shenghong Li bore contraction mappings in his mind when making this problem, and just combined the above picture with a Jensen's inequality (which requires that $a_1\leq\frac{1}{2}$ and $a_2>0$) to add to the difficulty of the problem.

We will prove the main claim: $ \left(\frac {n}{2(a_1 + a_2\dots + a_n)} - 1\right)^n \left(\frac {n}{a_1 + a_2 + \dots + a_n}\right)^n \leq \prod{ \left(\frac{1}{a_k}-1\right)} $. Let $f(x) = -\frac{(x-1)^2}{x-2}$ and $g(x) = \ln\left(\frac{1-x}{x}\right)$ where $0 \leq x \leq \frac{1}{2}$. First of all, we will prove $a_k \in \left(0, \frac{1}{2} \right]$ using induction; Since $f$ is a decreasing function, we have $f(0) = \frac{1}{2} \geq f(a_n) \geq f\left(\frac{1}{2}\right) > 0$ and thus we are done since $a_1$ clearly satisfies the given constraint. But then notice that $g$ is concave; since $g\left(\frac{m+n}{2}\right) \leq \frac{g(m) + g(n)}{2}$ where $m, n \in \left[0, \frac{1}{2} \right]$ implies $(m - n)^2 \geq 0$ by simple algebra, which is obviously true. Thus the main claim is proved and the finishing is trivial; $$\left(\frac{n}{\sum{a_k}}\right)^n \left(\frac{n}{2\sum{a_k}}-1 \right)^n \leq \left(\frac{\sum{(1-a_k)}}{\sum{a_k}} \right)^n \leq \prod{ \left(\frac{1}{a_k}-1\right)},$$where the first and second step is due to Cauchy and Jensen respectively.