Question

x,y∈R,x∧2+y∧2=4,求xy/(x+y-2)的最小值

Answer 1

x^2+y^2=4

(x+y)^2-2xy=4

(x+y)^2-4=2xy

(x+y-2)(x+y+2)=2xy

xy/(x+y-2)=(x+y+2)/2

设其=k

则x=2k-2-y

代入

(2k-2-y)^2+y^2=4

2y^2-2(2k-2)y+(2k-2)^2-4=0

关于y的方程

△=4(2k-2)^2-8[(2k-2)^2-4]≥0

(2k-2)^2≤8

1-√2≤k≤1+√2

所以xy/(x+y-2)的最小值为1-√2