设x>y>0,平面向量m=(x,1/x),n=(x,1/y+1/(x-y))则m·n的最小值
问题描述：
设x>y>0,平面向量m=(x,1/x),n=(x,1/y+1/(x-y))则m·n的最小值
答
mn=x^+(1/x)[1/y+1/(x-y)]=x^+1/(xy)+1/(x^-xy),设u=x^,v=xy,x>y>0,则u>v>0,mn=u+1/v+1/(u-v),记为f(u,v),u>v>0.下面用导数求驻点坐标：f'u=1-1/(u-v)^=0,f'v=-1/v^+1/(u-v)^=0,∴u-v=1,v=1,u=2.∴mn的最小值=f(2,1)...