Xét A=a33a−ab−ca+2bc:=a3a2+ab+ac−ab−ac+2bc
=a3a2+2bc
=a2.a+2abc−2abca2+2bc
=a−2abca2+2bc
Nên, S=a+b+c−2abc(1a2+2bc+1b2+2ca+1c2+2ab)+3abc
⩽
\leqslant 3 - 2abc + 3abc
= 3 + abc
\leqslant 3 + \frac{(a+b+c)^3}{27}
= 3+1 = 4
Dấu = xảy ra \Leftrightarrow a=b=c=1