Another nice Russian problem . We have these inequalities by AM-GM:$ \frac{1+x_{1}+x_{2}+...+x_{n}}{2}\ge\sqrt{(1+x_{1}+x_{2}+...+x_{n-1})\cdot x_{n}}$ so $ (1+x_{1}+x_{2}+...+x_{n})^{2}\ge 2^{2}(1+x_{1}+x_{2}+...+x_{n-1})x_{n}$. We have $ \frac{1+x_{1}+x_{2}+...+x_{n-1}}{3}\ge\sqrt [3]{\left(\frac{1+x_{1}+x_{2}+...+x_{n-2}}{2}\right)^{2}\cdot x_{n-1}}$ so $ (1+x_{1}+x_{2}+...+x_{n-1})^{3}\ge\frac{3^{3}}{2^{2}}(1+x_{1}+x_{2}+...+x_{n-2})^{2}x_{n-1}$. We have $ \frac{1+x_{1}+x_{2}+...+x_{n-2}}{4}\ge\sqrt [4]{\left(\frac{1+x_{1}+x_{2}+...+x_{n-3}}{3}\right)^{3}\cdot x_{n-2}}$ so $ ({1+x_{1}+x_{2}+...+x_{n-2})^{4}\ge\frac{4^{4}}{3^{3}}({1+x_{1}+x_{2}+...+x_{n-3})^{3}x_{n-2}}}$ $ \vdots$ $ (1+x_{1})^{n+1}\ge\frac{(n+1)^{n+1}}{n^{n}}x_{1}$. We multiply these and after we take $ \sqrt{\dots}$ both sides and we are done. Equality holds when $ x_{1}=\frac{1}{n}$ $ x_{2}=\frac{1+x_{1}}{n-1}$ $ x_{3}=\frac{1+x_{1}+x_{2}}{n-2}$ $ \vdots$ $ x_{n}= x_{1}+x_{2}+...x_{n-1}$

Wow, you are true! But why you can find this solution?

N.T.TUAN,do you know the other russian olympiad problems(this year)?Could you post it?

Erken wrote: N.T.TUAN,do you know the other russian olympiad problems(this year)?Could you post it? Yes , I know. Because i have got following book ''XXXIII Russian Mathematical Olympiad. Problems of the District and the Final rounds. 2006-2007 school year''. I'll post all them in National Olympiad box.

I have got all District and Final rounds for all grades for 1993-2006 years,nice book.But there isn't 2006-2007 olympiad. Is lorincz solution official?

These are two solutions in that book.

N.T.TUAN wrote: Wow, you are true! But why you can find this solution? I don't know what to say exactly, I've tried for case $ n=2$ and from it I had the idea. N.T.TUAN wrote: Erken wrote: N.T.TUAN,do you know the other russian olympiad problems(this year)?Could you post it? Yes , I know. Because i have got following book ''XXXIII Russian Mathematical Olympiad. Problems of the District and the Final rounds. 2006-2007 school year''. I'll post all them in National Olympiad box. I am eager to see them. Post them quickly N.T.TUAN. I'm wondering why are so few Russian problems posted in the Resources box.

lorincz...stud solution...The idea was awesome !!!

Erken wrote: I have got all District and Final rounds for all grades for 1993-2006 years,nice book.But there isn't 2006-2007 olympiad. Is lorincz solution official? Where can we find this book (what's its name?) It seems interesting

cancer wrote: Erken wrote: I have got all District and Final rounds for all grades for 1993-2006 years,nice book.But there isn't 2006-2007 olympiad. Is lorincz solution official? Where can we find this book (what's its name?) It seems interesting I don't know where you can free download it...But try to use google The name is "Всероссийские олимпиады школьников по математике 1993-2006",by Н.Агаханов. I bought this book when i was in Tuymaada 2007 olympiad... Good luck!

I also found it, interestingly. Here is my solution: Magical AM-GM: For $a_1;a_2;...;a_n\ge0 (n\in N^*)$. Prove that: $$(a^2_1+1)(a^2_1+a^2_2+1)....(a^2_1+a^2_2+a^2_3+...+a^2_n+1)\ge\sqrt{(n+1)^{n+1}}a_1a_2...a_n$$Solution: Denoting: $$x_1=\frac{a_1^2}{a_1^2+1}=1-\frac{1}{a_1^2+1}$$$$x_2=\frac{a_2^2}{(a_1^2+1)(a_1^2+a_2^2+1)}=\frac{1}{a_1^2+1}-\frac{1}{a_1^2+a_2^2+1}$$.... $$x_n=\frac{a_n^2}{(a_1^2+a_2^2+...+a_{n-1}^2+1)(a_1^2+a_2^2+...+a_{n-1}^2+a_n^2+1)}=\frac{1}{a_1^2+a_2^2+...+a_{n-1}^2+1}-\frac{1}{a_1^2+a_2^2+...+a_{n-1}^2+a_n^2+1}$$$$x_{n+1}=\frac{1}{a_1^2+a_2^2+...+a_{n-1}^2+a_n^2+1}$$$\implies x_1+x_2+...+x_n+x_{n+1}=1 $. Let : $$T=x_1x_2..x_nx_{n+1}=\left(\frac{a_1a_2...a_n}{(a^2_1+1)(a^2_1+a^2_2+1)....(a^2_1+a^2_2+a^2_3+...+a^2_n+1)}\right)^2$$Obviously by AM-GM inequality: $$T\leq\frac{1}{(n+1)^{n+1}}$$Thus, the inequality is proven. Equality holds iff: $$\frac{a_1^2}{a_1^2+1}=\frac{a_2^2}{(a_1^2+1)(a_1^2+a_2^2+1)}=...=\frac{a_n^2}{(a_1^2+a_2^2+...+a_{n-1}^2+1)(a_1^2+a_2^2+...+a_{n-1}^2+a_n^2+1)}=\frac{1}{a_1^2+a_2^2+...+a_{n-1}^2+a_n^2+1}=\frac{1}{n+1}$$