Theo AM-GM ta có:
$6= 3\frac{a^2}{3}+2\frac{b^2}{2}+c^2\geq 6\sqrt[6]{(\frac{a^2}{3})^3(\frac{b^2}{2})^2c^2}= 6\sqrt[6]{\frac{a^6b^4c^2}{3^32^2}}\Rightarrow a^6b^4c^2\leq 3^32^2$
Ta có:
$S=3\frac{a}{3bc}+4\frac{b}{2ca}+5\frac{c}{ab}\geq 12\sqrt[12]{(\frac{a}{bc})^3(\frac{b}{2ca})^4(\frac{c}{ab})^5}=\frac{12}{\sqrt[12]{3^32^4}}.\frac{1}{\sqrt[12]{a^6b^4c^2}}$
$\Rightarrow S\geq \frac{12}{\sqrt[12]{3^32^4}}.\frac{1}{\sqrt[12]{3^32^2}}=2\sqrt6$
Đẳng thức xảy ra $\Leftrightarrow a=\sqrt3,b=\sqrt2,c=1$