`0≤a≤b≤c≤1`
`⇔3≥a+b+c≥0`
ta có :
`a^2+b^2+c^2≥ab+bc+ac`
`⇔(a+b+c)^2≥3(ab+bc+ac)`
`⇔9≥(a+b+c)^2≥3(ab+bc+ac)`
`⇔3≥(ab+bc+ac)`
CM BĐTphụ
`a/(a+1)+b/(b+1)+c/(c+1)≥a+b+c-(ab^2)/(1+b^2)-(ca^2)/(1+b^2)-(bc^2)/(1+b^2)≥a+b+c-(ab+bc+ac)/2`
`B=(a+b+c+3)(1/(a+1)+1/(b+1)+1/(c+1))`
`⇔B=(a+b+c+3)(3-a/(a+1)-b/(b+1)-c/(c+1))`
`⇔B≤(a+b+c+3)(2(a+b+c)-(ab+bc+ac)/2)`
`⇔B≤(3+3)(2.3-3/2)`
`⇔B≤27`
`''=''`xẩy ra khi :
`a=b=c`
thay lại ko thỏa mãn
`⇒B` biểu thức `B` ko có `max`