I don't understand the problem Why is the first summation from i=1 to n and what the second sum means ?

mathangel wrote: I don't understand the problem Why is the first summation from i=1 to n and what the second sum means ? That's just a typo, i.e. the $ \sum_{i=1}^n b_k=1$ should be $ \sum_{k=1}^n b_k=1$ Others are correct.

Thank you for correcting my mistake, Dear shalex.

can you post a solution?

Is the answer M=3/2?

joao1 wrote: Is the answer M=3/2? Yeah!

Can any Chinese Mathlinkers post the solution of this problem?I waited it for a long time.

The proposer of this problem is Professor Li Weigu from Peking University. my solution: If all $ b_{i} (1\le i \le n)$ are given, it is trivial that $ a_{n}$ reaches its maxim al value iff all inequalities in condition (2) are equalities.In other word, w.l.o.g we can assume all inequalities in condition (2) are equalities. Compare the condition (2) of k and k-1 we get $ a_{k}^2-a_{k-1}^2=a_{k}b_{k}$ for k=0,1,…,n-1 (In order to make it convenient we let $ a_{0}$=1 Then we can get $ a_{k}=(b_{k}+\sqrt {b_{k}^2+4a_{k-1}^2})/2$ Since $ \sqrt {b_{k}^2+4a_{k-1}^2} > 2a_{k-1}$, we have $ a_{k}=(b_{k}+\sqrt {b_{k}^2+4a_{k-1}^2})/2 > (b_{k}+2a_{k-1})/2=a_{k-1}+\frac {b_{k}} {2}$. So $ a_{n} > a_{0}+\frac {\sum_{i=1}^{n} b_{i}} {2}=1+1/2=3/2$, which mean $ M=3/2$ satisfies the condition. On the other hand, since $ a_{i}$ is increasing and $ a_{1} > 1$,we have $ a_{i} > 1$, so $ a_{k-1}b_{k}^2 > b_{k}^2$ for every k. Thus $ \sqrt {b_{k}^2+4a_{k-1}^2} < \frac {b_{k}^2}{4}+2a_{k-1}$ and $ a_{k}=(b_{k}+\sqrt {b_{k}^2+4a_{k-1}^2})/2 < \frac {b_{k}+\frac {b_{k}^2}{4}+2a_{k-1}}{2}=a_{k-1}+\frac {b_{k}}{2}+\frac {b_{k}^2}{8}$ So $ a_{n}<a_{0}+\frac {\sum_{i=1}^{n} b_{i}} {2}+\frac {\sum_{i=1}^{n} b_{i}^2} {8}=3/2+\frac {\sum_{i=1}^{n} b_{i}^2}{8}$ Next we prove that $ \sum_{i=1}^{n} b_{i}^2$ can be arbitrarily small. Assume $ n=2m$(m is a positive integer).Since 2$ b_{k}\ge b_{k - 1} + b_{k + 1},k = 2,3,\cdots,n - 1$, it is easy to prove for $ 1<x<y<n$, $ b_{x-1}+b{y} \ge b_{x}+b{y-1}$.So for $ 1\le i \le m$, $ b_{i}+b_{2m} \le b_{i+1}+b_{2m-1} \le \cdots \le b_{[\frac {i+2m}{2}]-1}+b_{i+2m+1-[\frac {i+2m}{2}]}$. There are at least $ \frac {m-1}{2}$ terms in the former inequality links, so $ b_{i} < b_{i}+b_{2m} <\frac{\sum_{j=i}^2m b_j}{\frac {m-1}{2}}<\frac {1}{\frac {m-1}{2}=\frac {2}{m-1}$.i.e. for $ 1\le i \le m$, $ b_{i} < \frac {2}{m-1}$. For the same reason for $ m+1\le i \le 2m$, $ b_{i} < \frac {2}{m-1}$ Thus $ \sum_{i=1}^{n} b_{i}^2 < 2m(\frac {2}{m-1})^2=\frac {8m}{(m-1)^2}$, and for m is an arbitrary positive integer $ \frac {8m}{(m-1)^2}$ can be arbitrarily small. Combined with $ a_{n}<3/2+\frac {\sum_{i=1}^{n} b_{i}^2}{8}$ we get $ M$ can not be greater than 3/2.So the the maximal constant $ M$ is 3/2 Q.E.D PS: it pissed me for a long time because during the test I came up with this solution but failed for a extremely stupid reason. Right after I got $ a_{n}<3/2+\frac {\sum_{i=1}^{n} b_{i}^2}{8}$, of course I would guess whether $ \sum_{i=1}^{n} b_{i}^2$ can be arbitrarily small. I decided to take a set of special $ b_{i}$ to check it. Unfortunately I took $ b_{i}$ as a geometric progression which contradicts the condition (1)! And the result was that $ \sum_{i=1}^{n} b_{i}^2$ is about 1/3. This mistake cost me a lot of time and made me unable to guess the answer.Partly as a result I didn't make myself into the six members of the China team last year.

linboll wrote: The proposer of this problem is Professor Li Weigu from Peking University. my solution: If all $ b_{i} (1\le i \le n)$ are given, it is trivial that $ a_{n}$ reaches its maxim al value iff all inequalities in condition (2) are equalities.In other word, w.l.o.g we can assume all inequalities in condition (2) are equalities. Compare the condition (2) of k and k-1 we get $ a_{k}^2 - a_{k - 1}^2 = a_{k}b_{k}$ for k=0,1,…,n-1 (In order to make it convenient we let $ a_{0}$=1 Then we can get $ a_{k} = (b_{k} + \sqrt {b_{k}^2 + 4a_{k - 1}^2})/2$ Since $ \sqrt {b_{k}^2 + 4a_{k - 1}^2} > 2a_{k - 1}$, we have $ a_{k} = (b_{k} + \sqrt {b_{k}^2 + 4a_{k - 1}^2})/2 > (b_{k} + 2a_{k - 1})/2 = a_{k - 1} + \frac {b_{k}} {2}$. So $ a_{n} > a_{0} + \frac {\sum_{i = 1}^{n} b_{i}} {2} = 1 + 1/2 = 3/2$, which mean $ M = 3/2$ satisfies the condition. On the other hand, since $ a_{i}$ is increasing and $ a_{1} > 1$,we have $ a_{i} > 1$, so $ a_{k - 1}b_{k}^2 > b_{k}^2$ for every k. Thus $ \sqrt {b_{k}^2 + 4a_{k - 1}^2} < \frac {b_{k}^2}{4} + 2a_{k - 1}$ and $ a_{k} = (b_{k} + \sqrt {b_{k}^2 + 4a_{k - 1}^2})/2 < \frac {b_{k} + \frac {b_{k}^2}{4} + 2a_{k - 1}}{2} = a_{k - 1} + \frac {b_{k}}{2} + \frac {b_{k}^2}{8}$ So $ a_{n} < a_{0} + \frac {\sum_{i = 1}^{n} b_{i}} {2} + \frac {\sum_{i = 1}^{n} b_{i}^2} {8} = 3/2 + \frac {\sum_{i = 1}^{n} b_{i}^2}{8}$ Next we prove that $ \sum_{i = 1}^{n} b_{i}^2$ can be arbitrarily small. Assume $ n = 2m$(m is a positive integer).Since 2$ b_{k} \ge b_{k - 1} + b_{k + 1},k = 2,3,\cdots,n - 1$, it is easy to prove for $ 1 < x < y < n$, $ b_{x - 1} + b_{y} \ge b_{x} + b_{y - 1}$.So for $ 1\le i \le m$, $ b_{i} + b_{2m} \le b_{i + 1} + b_{2m - 1} \le \cdots \le b_{[\frac {i + 2m}{2}] - 1} + b_{i + 2m + 1 - [\frac {i + 2m}{2}]}$. There are at least $ \frac {m - 1}{2}$ terms in the former inequality links, so $ b_{i} < b_{i} + b_{2m} < \frac{ \sum_{j = i}^{2m} b_j }{ \frac{m - 1}{2} } < \frac{1}{\frac{m - 1}{2}} = \frac{2}{m - 1}$.i.e. for $ 1\le i \le m$, $ b_{i} < \frac {2}{m - 1}$. For the same reason for $ m + 1\le i \le 2m$, $ b_{i} < \frac {2}{m - 1}$ Thus $ \sum_{i = 1}^{n} b_{i}^2 < 2m(\frac {2}{m - 1})^2 = \frac {8m}{(m - 1)^2}$, and for m is an arbitrary positive integer $ \frac {8m}{(m - 1)^2}$ can be arbitrarily small. Combined with $ a_{n} < 3/2 + \frac {\sum_{i = 1}^{n} b_{i}^2}{8}$ we get $ M$ can not be greater than 3/2.So the the maximal constant $ M$ is 3/2 Q.E.D PS: it pissed me for a long time because during the test I came up with this solution but failed for a extremely stupid reason. Right after I got $ a_{n} < 3/2 + \frac {\sum_{i = 1}^{n} b_{i}^2}{8}$, of course I would guess whether $ \sum_{i = 1}^{n} b_{i}^2$ can be arbitrarily small. I decided to take a set of special $ b_{i}$ to check it. Unfortunately I took $ b_{i}$ as a geometric progression which contradicts the condition (1)! And the result was that $ \sum_{i = 1}^{n} b_{i}^2$ is about 1/3. This mistake cost me a lot of time and made me unable to guess the answer.Partly as a result I didn't make myself into the six members of the China team last year.

Fantastic! Fantastic!

Let me give another approach to bound $a_n$ from above: as linboll pointed out we may assume $a_k^2-a_{k-1}^2=a_kb_k$ for $1\leq k \leq n$ (where $a_0=1$), that is \[a_1-1=\frac{a_1}{a_1+1}b_1,a_2-a_1=\frac{a_2}{a_2+a_1}b_2,...,a_n-a_{n-1}=\frac{a_n}{a_n+a_{n-1}}b_n.\]Also it's not hard to show that (1) implies $b_k\leq\frac{2}{n-1}(1\leq k \leq n)$. Then we have \[2a_n-3=\sum_{k=1}^n[2(a_k-a_{k-1})-b_k]=\sum_{k=1}^n\frac{a_k-a_{k-1}}{a_k+a_{k-1}}b_k\leq\sum_{k=1}^n\frac{a_k-a_{k-1}}{n-1}=\frac{a_n-1}{n-1},\]i.e. $a_n\leq\frac{3}{2}+\frac{1}{4n-6}$. Thus $M\leq\frac{3}{2}$.