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芝士点¶
轮换和(Cyclic Sum): 考虑一个函数\(f(x_1,x_2,\cdots,x_n)\)，它的轮换和定义为
\(\displaystyle\sum _{cyc}f(x_1,x_2,\cdots,x_n)=f(x_1,x_2,\cdots,x_{n-1},x_n)+f(x_2,x_3\cdots,x_n,x_1)+\cdots +f(x_n,x_1,\cdots,x_{n-2},x_{n-1})\)
例题¶
正实数x,y,z满足\(xyz\ge1\)，证明\(\displaystyle\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5} \ge 0\)
解：
\(\because\) \(\displaystyle\frac{x^5-x^2}{x^5+y^2+z^2}-\frac{x^5-x^2}{x^3(x^5+y^2+z^2)}=\frac{x^5-x^2}{x^3(x^5+y^2+z^2)(x^2+y^2+z^2)}\ge0\)
\(\therefore\) \(\displaystyle\sum _{cyc}\frac{x^5-x^2}{x^5+y^2+z^2}\ge\sum _{cyc}\frac{x^5-x^2}{x^3(x^2+y^2+z^2)}=\frac{1}{x^2+y^2+z^2}\sum _{cyc}(x^2-\frac{1}{x})\ge\frac{1}{x^2+y^2+z^2}\sum _{cyc}(x^2-yz)\ge0\)
得证
题源：2005-46th-IMO-3