Áp dụng bất đẳng thức bunhiacopski dạng phân thức(cosi-schwars) ta có:
`1/x+1/x+1/y+1/z>=(1+1+1+1)^2/(2x+y+z)=16/(2x+y+z)`
`<=>1/(2x+y+z)<=1/16(1/x+1/x+1/y+1/z)`
Hoàn toàn tương tự ta có:
`1/(x+2y+z)<=1/16(1/x+1/y+1/y+1/z)`
`1/(x+y+2z)<=1/16(1/x+1/y+1/z+1/z)`
`=>1/(2x+y+z)+1/(x+2y+z)+1/(x+y+2z)<=1/16(4/x+4/y+4/z)`
`<=>1/(2x+y+z)+1/(x+2y+z)+1/(x+y+2z)<=1/16*16=1(đpcm)`
Dấu "=" xảy ra khi `x=y=z=3/4.`