Please post your solutions. Use $\LaTeX$ please! Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

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no one have solution?

solution

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hxy09 wrote: solution nice solution

hxy09 wrote: solution A fomous university in China... The paper is your examination test paper...

Jianxing113725 wrote: hxy09 wrote: solution A fomous university in China... The paper is your examination test paper... Very sorry,the paper isn't mine,my mother is a teacher in this university.She gave me these papers to write my solution down I am only a high school student

Anyone have different ideas?Solution posted by hxy09 is the official solution...It's unimaginable I think....

We will prove the result by strong induction on $k$. When $k = 1$, the result is obvious. Now, suppose the result holds for $k < t.$ When $k = t,$ we consider two cases. Case 1. There exists an $1 \le i \le k-1$ such that $n_1 n_2 \cdots n_i \le (4a)^{2^i-1}.$ Then, observe that: $$\prod_{j = i+1}^{k} \left(1 - \frac{1}{n_j}\right) \le \frac{a n_1 n_2 \cdots n_i}{b (n_1 - 1)(n_2-1)\cdots (n_i-1)} < \prod_{j=i+1}^{k-1} \left(1- \frac{1}{n_j}\right).$$Hence, since $k - i \ge 1$, the inductive hypothesis implies that: $$(4an_1 n_2 \cdots n_i)^{2^{k-i}-1} \ge \prod_{j= i+1}^{k} n_j. \qquad (1)$$Therefore, observe that from our assumption that $(4a)^{2^i-1} \ge n_1n_2 \cdots n_i$, we get: $$((4a)^{2^i})^{2^{k-i}-1} \cdot n_1n_2 \cdots n_i \ge \left(\prod_{j=i+1}^{k} n_j\right) \cdot n_1n_2 \cdots n_i,$$so that $$(4a)^{2^k-1} \ge ((4a)^{2^i})^{2^{k-i}-1} \cdot n_1n_2 \cdots n_i \ge \prod_{i=1}^{k} n_i,$$as desired. Case 2. There does not exist an $1 \le i \le k-1$ such that $n_1n_2 \cdots n_i \le (4a)^{2^i-1}.$ Then, from the fact that $n_1 < n_2 < \cdots < n_k$ we get that: $$n_i > \left ((4a)^{2^i-1} \right)^{\frac1i}, \forall 1 \le i \le k-2.$$Since $\frac{2^i-1}{i} \ge i-1$ for $i \ge 2$, we get that $n_1 > 4a$ and $n_i > (4a)^{i-1}$ for $2 \le i \le k-1.$ We also clearly have $n_k > n_{k-1} > (4a)^{k-2}.$ For $k \le 4$, this already yields from $4a < n_1 < n_2 < \cdots < n_k$ and telescoping that: $$\prod_{i=1}^{k} \left( 1 - \frac{1}{n_i} \right) \ge \frac{4a+1}{4a+k+1} < \frac{a}{a+1} < \frac{a}{b},$$which is clearly absurd. Hence, we can assume that $k \ge 5$ from now on. Observe from telescoping that: $$\left(1-\frac{1}{4ax-4a+1}\right)\left(1 - \frac{1}{4ax-4a+2}\right) \cdots \left( 1 - \frac{1}{4ax}\right) = \frac{x-1}{x},$$ and so since $1 - \frac{1}{4ax}$ is greater than or equal to all of the multiplicands on the LHS we obtain that $$1 - \frac{1}{4ax} \ge \left(1 - \frac{1}{x}\right)^{\frac1{4a}}, \forall x \in \mathbb{N}.$$A simple induction then yields that: $$1 - \frac{1}{(4a)^{i}} \ge \left(1 - \frac{1}{4a}\right)^{\frac{1}{(4a)^{i-1}}}, \forall i \in \mathbb{N}$$This yields for $k \ge 5$ that: $$\prod_{i=1}^{k} \left(1 - \frac1{n_i}\right) \ge \left (1 - \frac1{4a}\right )^{\frac{1}{(4a)^0} + \frac{1}{(4a)^0} + \frac{1}{(4a)^1} + \frac{1}{(4a)^2} \cdots + \frac{1}{(4a)^{k-4}} + \frac{1}{(4a)^{k-3}} + \frac{1}{(4a)^{k-3}}}.$$For $k \ge 5$, it's easy to verify that the exponent is at most $2.5$. Therefore, we have that if $a \ge 2$, then: $$\prod_{i=1}^{k} \left(1 - \frac1{n_i}\right) \ge \left(1-\frac{1}{4a}\right)^{2.5} > \frac{a}{a+1} \geq \frac{a}{b},$$where the second inequality used above is easily verified. This is a clear absurdity. It remains only to tackle the case where $a = 1.$ To do so, we will use the improvements $1 - \frac{1}{n_1} \ge 1 - \frac{1}{4a+1} = \frac{4}{5}$ and $1 - \frac{1}{n_2} \ge 1 - \frac{1}{4a+2} = \frac{5}{6}.$ Implementing these then yields the stronger bound of: $$\prod_{i = 1}^{k} \left(1 - \frac{1}{n_i}\right) \ge \frac23 \cdot (1 - \frac14)^{0.5} > \frac12 \ge \frac{a}{b}.$$This is clearly absurd. As we've reached absurdity in all cases, we in fact obtain that Case $2$ is not even possible. Hence, the induction is complete, and the problem is solved. $\square$

Sorry but shouldnt this problem be in chinese if it is a china test

Actually the best bound $\frac{(a+1)^{2^k}-(a+1)^{2^{k-1}}}{a}$ can be worked out by induction+Karamata Inequality(I should have post this some times earlier, but I finally found the source today)

Translating @ETHANWYX2009 proof in PNG to latex using chatgpt : Let positive integers $a_1 < a_2 < \dots < a_n$ and $a, b \in \mathbb{N}$ such that: \[ \frac{(1 - \frac{a_1}{a})^2}{a_1} \cdot \frac{(1 - \frac{a_2}{a})^2}{a_2} \cdots \frac{(1 - \frac{a_n}{a})^2}{a_n} < \frac{(1 - \frac{a_1}{b})^2}{a_1} \cdot \frac{(1 - \frac{a_2}{b})^2}{a_2} \cdots \frac{(1 - \frac{a_n}{b})^2}{a_n} \] Proof: $a_1 \cdots a_n \leq \frac{(a+1)^2 - (a+n)^2}{a}$. Solution: Induction on $n$. The base case is trivial. Assume the inequality holds for $k < n$. Consider $k = n$. By the induction hypothesis, we have: \[ a_1 \cdots a_n > \frac{(a+1)^2 - (a+n)^2}{a} \]Then, by condition: \[ \frac{(1 - \frac{a_1}{a}) \cdots (1 - \frac{a_n}{a})}{b \cdot (a_1 \cdots a_n)} \leq \frac{a_1 \cdots a_n}{a} < \frac{(1 - \frac{a_1}{b})^2}{a_1} \cdots \frac{(1 - \frac{a_n}{b})^2}{a_n} \]Using the induction hypothesis, we get: \[ a_1 \cdots a_n \leq \frac{(a_{i+1} - a_{i})^2}{a_1} \cdots \frac{(a_{n+1} - a_{n})^2}{a_n} < \frac{(a+1)^2 - (a+i+1)}{a \cdot a_1 \cdots a_n} \]Therefore, $a_1 \cdots a_n \leq \frac{(a+1)^2}{a_1} \cdots \frac{(a_{n+1} - a_{n})^2}{a_n}$. Let $t_i = \frac{(a+1)^2}{a_1} \cdots \frac{(a_{n+1} - a_{n})^2}{a_n}$. For $i = 1, 2, \dots, n$. We prove the following lemma: \[ a_n \cdots a_2 = t_2 \cdots t_n, \quad \forall k \in \{1, 2, \dots, n\} \] Proof: If $a_n - a_i < t_1 - t_i = \frac{(a+1)^2}{a}$, then: \[ a_1 \cdots a_i < \frac{(a+1)^2}{a_1} \cdots \frac{(a_{n+1} - a_{n})^2}{a_n} \Rightarrow a_1 \cdots a_n \leq \frac{(a+1)^2}{a} \]which is a contradiction! Thus, $\sum_{i=1}^n t_i = t_1 + t_2 + \cdots + t_n$. Construct $f(x) = P(x) \cdot \ln (e^x)$. Then $f'(x) < 0$. By Karamata's inequality: \[ f(\ln t_1) + \cdots + f(\ln t_n) \geq f(\ln t_2) + \cdots + f(\ln t_n) \]This implies that: \[ (1 - \frac{a_1}{a}) \cdots (1 - \frac{a_n}{a}) \geq (1 - \frac{t_1}{t_1}) \cdots (1 - \frac{t_n}{t_n}) \]and that: \[ \frac{(a+1)^2 - t_n t_n^{-1}}{a} t_n > \frac{(a+1)^2}{a} \Rightarrow t_n \cdots t_n^{-1} = \frac{a_1}{a_n - 1} = \frac{a_{n+1}}{a_n + 1} \Rightarrow t_n = t_n - 1 \]However, $(1 - \frac{a_1}{a}) \cdots (1 - \frac{t_n}{t_n}) = \frac{a_1 \cdots a_n}{t_1 \cdots t_n}$, which is a contradiction!