Vì `x,y,z>0,` áp dụng BĐT Côsi ta có:
`x+y+z>=3\root{3}{xyz}`
`=>\root{3}{xyz}<=1/3`
`=>xyz<=1/27`
Áp dụng BĐT Côsi cho `x+y;y+z;x+z` ta có:
`<=> (x+y)(y+z)(x+z)=2(x+y+z)`
`(x+y)(y+z)(x+z)>=3\root{3}{(x+y)(y+z)(x+z)}`
`=>2>=3\root{3}{(x+y)(y+z)(x+z)}`
Dấu "=" xãy ra khi
`x=y=z=1/3=>S<=8/27 . 1/27=8/729`
Vậy `S_max=8/729` khi `x=y=z=1/3.`