Đặt \(\left\{{}\begin{matrix}x=\sqrt{3}a\\y=\sqrt{3}b\\z=\sqrt{3}c\end{matrix}\right.\Rightarrow x^2+y^2+z^2=3\)
Áp dụng bất đẳng thức Cauchy, ta có:
\(3=x^2+y^2+z^2\ge3\sqrt[3]{\left(xyz\right)^2}\Rightarrow xyz\le1\)
\(S=\dfrac{1}{\sqrt{3}}\left(x+y+z\right)+\dfrac{3\sqrt{3}}{xyz}+\dfrac{1}{3}\sum\dfrac{x^4}{yz}\)
\(\ge4\sqrt[4]{\dfrac{1}{\sqrt{3}}.\left(x+y+z\right).\left(\dfrac{\sqrt{3}}{xyz}\right)^3}+\dfrac{1}{3}.\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+zx}\)
\(\ge4\sqrt[4]{\dfrac{1}{\sqrt{3}}.3\sqrt[3]{xyz}.\dfrac{3\sqrt{3}}{\left(xyz\right)^3}}+\dfrac{1}{3}.\dfrac{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}{xy+yz+zx}\)
\(=4\sqrt[4]{\dfrac{9}{\sqrt[3]{\left(xyz\right)^8}}}+\dfrac{1}{3}.\dfrac{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}{xy+yz+zx}\)
\(\ge4\sqrt[4]{9}+1=4\sqrt{3}+1\)
Vậy \(MinS=4\sqrt{3}+1\), đạt được tại \(a=b=c=\dfrac{\sqrt{3}}{3}\)
