Question

若正数x,y,z满足xyz(x+y+z)=4,则(x+y)(y+z)的最小值为多少?

Answer 1

因为x,y,z>0,则:

(x+y)(y+z)=xy+yy+xz+yz=(x+y+z)y+xz,令:

(x+y+z)y=a,xz=b,则:

(a+b)/2≥(ab)^(1/2)=((x+y+z)xyz)^(1/2)=4^(1/2)=2,

故(a+b)min=4,所以(x+y)(y+z)min=(a+b)min=4,最小值为4.

如果我没算错,应该是这样.