TH1 : `x≤0`
`=>x/((x+2004)^2)<0<1/8016 forall x`
Th2 : ` x >0`
Đặt `A=x/(x+2004)^2`
`=>1/A=(x+2004)^2/x`
`=>1/A=(x^2+2.2004x+2004^2)/x`
`=>1/A=(4.2004x+x^2-2.2004+2004^2)/x`
`=>1/A=8016+((x-2004)^2)/x≥8016 forall x`
`=>A≤8016 forall x`
`=>x/(x+2004)^2 ≤8016 forall x`
Dấu "=" xảy ra `<=>x=2004`