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đặt $x=a+b, y=a+c, z=b+c\Rightarrow x+2y+3z=12$ $\frac{ab}{ab+a+b}=1-\frac{a+b}{ab+a+b}\leq 1-\frac{a+b}{\frac{(a+b)^2}{4}+a+b}=1-\frac{4}{x+4}\leq \frac{x+1}{9}$ tương tự: $\frac{2ac}{ac+a+c}\leq 2-\frac{8}{y+4}\leq \frac{2y+2}{9}$ $\frac{3bc}{bc+b+c}\leq \frac{3z+3}{9}$ suy ra $S\leq 2$ Dấu bằng xảy ra khi $a=b=c=1$
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