Question

试分解正数a为三个正数之和，并使他们的倒数和为最小

Answer 1

x+y+z=a

x+y+z>=3(xyz)^(1/3)=3(1/(1/(xyz)^(1/3)))

>=3*3(1/((1/x)+(1/y)+(1/z)))

所以:(1/x)+(1/y)+(1/z)>=9(1/(x+y+z))=9/a

当x=y=z=a/3时,他们的倒数和为最小=9/a