已知x,y属于R+,x+y=xy,求u=x+2y的最小值
问题描述：
已知x,y属于R+,x+y=xy,求u=x+2y的最小值
答
x+y=xy
(y-1)x=y
x=y/(y-1)>0
又y>0,因此y-1>0 y>1
u=x+2y
=y/(y-1)+2y
=[y+2y(y-1)]/(y-1)
=(2y^2-y)/(y-1)
=(2y^2-2y+y-1+1)/(y-1)
=2y+1+1/(y-1)
=2(y-1)+1/(y-1)+3
由均值不等式,得
2(y-1)=1/(y-1)时,u取到最小值.
2(y-1)=1/(y-1)
(y-1)^2=1/2
y-1=√2/2
y=1+√2/2
此时,u有最小值umin=3+2√2