设x y∈R x²＋y²=4 则2xy/(x+y-2)的最小值是

题目

设x y∈R x²＋y²=4 则2xy/(x+y-2)的最小值是

答案

已知x²+y²=4,求2xy/(x+y-2)的最小值.
由于(x-y)²≥0,展开得：2xy≤x²+y²,则有：
x²+y²+2xy≤2(x²+y²)
(x+y)²≤2(x²+y²)=8
得：-2√2≤x+y≤2√2,
所以有：
2xy/(x+y-2)
=(x²+y²+2xy-4)/(x+y-2)
=[(x+y)²-4]/(x+y-2)
=(x+y+2)(x+y-2)/(x+y-2)
=x+y+2≥2-2√2
因此,2xy/(x+y-2)的最小值是2-2√2.