设x,y属于r.且x^+y^=4,则2xy/x+y-2的最小值
问题描述：
设x,y属于r.且x^+y^=4,则2xy/x+y-2的最小值
答
2xy/(x+y-2)=[(x+y)^2-(x^2+y^2)]/(x+y-2)又x^2+y^2＝4
所以=[(x+y)^2-4]/(x+y-2)
=[(x+y-2)(x+y+2)]/(x+y-2)
=x+y+2
又因为x^2+y^2>=(x+y)^2/2
|x+y|>=根号[2(x^2+y^2)]=2根号2
所以x+y+2>=2-2根号2
Min＝2－2根号2