S=√x+√y+√z
S²=x+y+z+2(√xy+√yz+√zx)
因x,y,z≥0,则(√xy+√yz+√zx)≥0
则S²≥x+y+z=1
=>S≥1
又2√xy≤x+y,2√yz≤y+z,2√zx≤x+z
则S²≤3(x+y+z)=3
=>1≤S≤√3