$4c+2b\geq a(b^{2}+c^{2})\geq 2abc\Rightarrow \frac{2}{b} +\frac{1}{c}\geq a \Rightarrow S=\left ( \frac{1}{b+c-a}+\frac{1}{a+c-b} \right )+2\left ( \frac{1}{b+c-a}+\frac{1}{a+b-c} \right )+3\left ( \frac{1}{a+c-b}+\frac{1}{a+b-c}\right )\geq \frac{2}{c}+\frac{4}{b}+\frac{6}{a} \geq 2a+\frac{6}{a}\geq 4\sqrt{3}$
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$