摘要：设 $n\in N,n\ge2,a>0$ 证明 :$(a-1)((n+1)a^n+(n-1)-(n+1)a-(n-1)a^{n+1})\le0$ 阅读全文

摘要：设$a_i\in R^{*}(i=1,2,...n)$,求证：$\frac{1}{a_1}+\frac{2}{a_1+a_2}+\cdots+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}}\le 2\sum_{i=1}^{n}\frac{1}{a_i}$ 阅读全文

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摘要：设$a_{i}>0,b_{i}>0,i=1,2,\cdots ,n.$求证：$$\sum_{i=1}^{n}\frac{a_{i}b_{i}}{a_{i}+b{i}}\le \frac{(\sum_{i=1}^{n}a_{i})(\sum_{i=1}^{n}b_{i})}{\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i}}$$见<<数学奥林匹克教程>>P15 阅读全文

摘要：已知$n\in N^{*}$ 若$n^4+2n^3+5n^2+12n+5$为完全平方数，则$n=_____$Solve:设$m^2=n^4+2n^3+5n^2+12n+5,$$\because (n^2+n+3)^2\ge (n^2+n+3)^2-2(n-1)(n-2)= m^2=(n^2+n+2)^2+8n+1\ge (n^2+n+2)^2,$$\therefore m^2=(n^2+n+2)^2 orm^2=(n^2+n+3)^2$ 容易验证当$m^2=(n^2+n+3)^2$时，$n=1 or n=2$ 阅读全文

摘要：\[ \left\{\begin{matrix}x^4+y^2-xy^3-\tfrac{9}{8}x=0 &\\ y^4+x^2-yx^3-\tfrac{9}{8}y=0 &\end{matrix}\right. \]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=104479 阅读全文

摘要：$ a^3+b^3+c^3-a^2b-b^2c-c^2a=\frac{\sum_{cyc}(3a^2-2ab-c^2)^2+(\sum_{cyc}(a-b)^2)((a+b+c)^2+5(ab+bc+ca))}{12(a+b+c)} $http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=406991&p=2273149#p2273149 阅读全文