Xét $A=\frac{a^3}{3a-ab-ca+2bc}$:$= \frac{a^3}{a^2+ab+ac-ab-ac+2bc}$$= \frac{a^3}{a^2+2bc}$$= \frac{a^2.a+2abc-2abc}{a^2+2bc}$ $= a- \frac{2abc}{a^2+2bc}$Nên, $S=a+b+c-2abc(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab})+3abc$ $\leqslant - 2abc (\frac{9}{(a+b+c)^2}) +3abc$ $\leqslant 3 - 2abc + 3abc $ $= 3 + abc$ $\leqslant 3 + \frac{(a+b+c)^3}{27}$ $= 3+1 = 4$Dấu $=$ xảy ra $\Leftrightarrow$ $a=b=c=1$

Xét $A=\frac{a^3}{3a-ab-ca+2bc}$:$= \frac{a^3}{a^2+ab+ac-ab-ac+2bc}$$= \frac{a^3}{a^2+2bc}$$= \frac{a^2.a+2abc-2abc}{a^2+2bc}$ $= a- \frac{2abc}{a^2+2bc}$Nên, $S=a+b+c-2abc(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab})+3abc$ $\leqslant 3- 2abc (\frac{9}{(a+b+c)^2}) +3abc$ $\leqslant 3 - 2abc + 3abc $ $= 3 + abc$ $\leqslant 3 + \frac{(a+b+c)^3}{27}$ $= 3+1 = 4$Dấu $=$ xảy ra $\Leftrightarrow$ $a=b=c=1$