Question

已知x,y属于R+,x+y=xy,求u=x+2y的最小值

Answer 1

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