I think this problem can be solved by using Induction method, but the given condition is: $ a_{1}\le b_{i}\le 2a_{i}, i=1,2,\ldots,k $ or $ a_{i}\le b_{i}\le 2a_{i}, i=1,2,\ldots,k $

Yes, it should be $a_i\leq b_i\leq 2a_i$.

This is the remain unsolved problem of China TST 2012. Can anyone give a solution or idea of this?

I think the problem can't be solved by induction because the property of each case of $k$. With each set of number such that $a_i \le b_i \le 2a_i$, we can choose $2a_i-1$ case of $b_i$ as $a_i,a_i+1,a_i+2,...,2a_i$ and with each case, we can choose some of $c_i=0$ to hold the first inequality. There are $(2a_1-1)(2a_2-1)...(2a_k-1)$ case of $(b_1,b_2,...,b_k)$. Hence, to hold the second inequality, each of $a_i$ must be greater than $k$ and if we assume $a_1<a_2<...<a_k$ then $a_1 > k, a_2 > ka_1,a_3>ka_2,...,a_{k}>ka_{k-1}$. I just get a little result, can any one solve it?

The official solution is by induction. First generalize the problem as follows: for any positive real number $k$ and any positive interger $n$, there exist $n$ pairwise distinct positive integers $a_1,a_2,\cdots,a_n$, such that (i) $2a_i<a_{i+1}$, $i=1,2,\cdots,n-1$; (ii) for any real numbers $b_i\in [a_i,2a_i]$, and non-negative integers $c_i$, $i=1,2,\cdots,n$, if $\prod_{i=1}^n b_i^{c_i}<\prod_{i=1}^n b_i$, then $k\prod_{i=1}^n b_i^{c_i}<\prod_{i=1}^n b_i$. Then induct on $n$. The case for $n=1$ is simple. Suppose $x_1,x_2,\cdots,x_n$ is an example for the case of $n$, then choose $x_{n+1}>2x_n$, satisfying (1) $\frac{x_{n+1}}{2x_n}>k\left(\frac{2x_n}{x_1}\right)^n$. Let $a_i=tx_i$, $i=1,2,\cdots,n+1$, $t$ is large enough so that (2) $a_1^{n+1}>2^n a_2\cdots a_{n+1}$; (3) $k\cdot 2^{n-1} a_{n+1}^{n-1}<a_1a_2\cdots a_n$. Now using the inequalities (1)(2) and (3), it's not hard to show that $a_1,a_2,\cdots,a_{n+1}$ has the required property for the case of $n+1$. (One shall discuss several cases for $c_i$'s).

let x be a positive integer seq that grows fast enough let X much bigger than x[n] let a=X*x

The idea is quite easy, but hard to think of. Choose $n$ integers such that the lowest possible number raised to the $n+1$ will exceed the product of the highest possible value of the other elements. The way we do this is by scaling each element, since one term has degree $n+1$ and the product is just multiplied by a constant. The upshot of this is that the maximal way to arrange the numbers in $c_i$ is $n-1$ of the largest element and $1$ of the smallest element, because $n$ of the largest element is obviously too big. However, if the ratio of the largest to the second largest is large enough, this property will be met.

Solution: We start with the following lemma. Lemma: For any positive integer $l$, there exists $l$ distinct positive integers $d_1,\dots,d_l$, such that the only solution of the equation $$\sum\limits_{i=1}^{l}x_{i}d_{i}=0$$in $(\mathbb{Z}\cap (-\infty,1])^l$ is $(0,0,\dots,0).$ Proof of the lemma: Induct on $l$. The case $l=1$ is trivial. Suppose that the lemma holds for $l-1$, and $d_1',\dots,d_{l-1}'$ satisfies the condition. Take $d_{i}=2d_{i}'$ for $1\leq i\leq l-1$, and take $d_{l}$ to be an odd integer larger than $2\sum\limits_{i=1}^{l-1}d_{i}$. Notice that if $\sum\limits_{i=1}^{l}x_id_i=0$, we must have $x_{l}$ is even, then $x_l$ is nonpositive. If $x_l$ is less than $0$, then clearly $\sum\limits_{i=1}^{l}x_id_i<0$, contradiction. Thus $x_l=0$. By the inductive hypothesis we conclude that the lemma holds for $l$ as well. Therefore the lemma is proved. $\blacksquare$ Back to the main problem, consider any positive integer $N$. Take $d_1,\dots,d_k$ in the lemma, and set $a_i=2^{Ad_i}$ for $A>N+k+\sum\limits_{i,j}\frac{d_i}{d_j}$. Then we know that $\text{log}_2b_i\in [Ad_i,Ad_i+1]$. Denote the difference $\text{log}_2b_i-Ad_i$ by $\delta_i$ . From the condition we know that $\sum\limits_{i=1}^k x_i(Ad_i+\delta_i)>0$ for some integers $x_i=1-c_i\leq 1$. Because $\sum\limits_{i=1}^k x_i\delta_i\leq k$, we know that $\sum\limits_{i=1}^k x_id_i\geq0$, therefore it is at least $1$. Note that $d_jx_j+\sum\limits_{i\neq j}d_i\geq \sum\limits_{i=1}^kx_id_i>0,$, then $x_j>-\sum\limits_{i\neq j}\frac{d_i}{d_j}$. As a result, $$\sum\limits_{i=1}^k x_i\text{log}_2b_i=A\sum\limits_{i=1}^kx_id_i+\sum\limits_{i=1}^kx_i\delta_i\geq A+\sum\limits_{x_i<0}x_i >N$$This shows that $$2^N\prod_{i=1}^{k}b_i^{c_i}<\prod_{i=1}^{k}b_i.$$where $N$ can be taken arbitrarily large. And we are done!

can it be done by using abel summation ? Anyone @above