Question

设x,y,z≥0,x+y+z=3,证明:√x+√y+√z≥xy+yz+zx

Answer 1

平方用“方”代替 =大于等于

等价于证2√x+2√y+2√z>=2xy+2yz+2zx

.x(3-x)+y(3-y)+z(3-z)

3(x+y+z)-x方-y方-z方

9-x方-y方-z方

3+3+(x+y+z)-x方-y方-z方

-(x-2√x+1）-(y-2√y+1)-(z-2√z+1)>=-x方-y方-z方+3

(√x-1)方+（√y-1）方+（√z-1）方0

所以0