Perhaps this is a simple matter. an application by a simple format of equality Abel. I really want to have a solution in 2003.2004(Chinese also). here is an image.

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math10 wrote: Perhaps this is a simple matter. an application by a simple format of equality Abel. I really want to have a solution in 2003.2004(Chinese also). here is an image. As I can not understand the language(either its Chinese or Japanese(kanji), please post your solution in english. Remark: The scanning was unclear.! I will try understanding it after you or any other member translates it..

What I think it's saying: The necessary and sufficient condition is that there is some $ k$ for which $ \sum_{i=k}^n x_i>0$. For necessity, note that we can write \[ \sum a_ix_i = x_1(a_1+\dots+a_n)+(x_2-x_1)(a_2+\dots+a_n)+(x_3-x_2)(a_3+\dots+a_n)+\dots+(x_n-x_{n-1})a_n.\] Since the differences of the $ x_i$ are by assumption positive, at least one of the sums of the $ a_i$ must be positive. For sufficiency, let us suppose that the statement holds for some $ k$. Then let $ x_1=1, x_2=2, \dots x_{k-1}=k-1, x_k=m+k, x_{k+1}=m+k+1, \dots x_n=m+k+n$, where $ m$ is to be determined later. Then $ \sum a_ix_i$ has the form $ m(x_{k}+x_{k+1}+\dots+x_n)+c$, where $ c$ is some constant not depending on $ m$. For large enough $ m$, this is positive.