Áp dụng BĐT Bunhiacopski(mở rộng):
$(\sqrt[4]{a^{3}}+\sqrt[4]{b^{3}}+\sqrt[4]{c^{3}})^{4}$
$=(\sqrt[4]{\frac{a}{x}}.\sqrt[4]{\frac{a}{x}}.\sqrt[4]{\frac{a}{x}}.\sqrt[4]{x^{3}}+\sqrt[4]{\frac{b}{y}}.\sqrt[4]{\frac{b}{y}}.\sqrt[4]{\frac{b}{y}}.\sqrt[4]{y^{3}}+$
$+\sqrt[4]{\frac{c}{z}}.\sqrt[4]{\frac{c}{z}}.\sqrt[4]{\frac{c}{z}}.\sqrt[4]{z^{3}})^{4}$
$\leq (\frac{a}{x}+\frac{b}{y}+\frac{c}{z})^{3}(x^{3}+y^{3}+z^{3})=S$
$\Rightarrow S\geq (\sqrt[4]{a^{3}}+\sqrt[4]{b^{3}}+\sqrt[4]{c^{3}})^{4}$
Dấu "=" xảy ra $\Leftrightarrow \begin{cases}\frac{\frac{a}{x}}{x^{3}}=\frac{\frac{b}{y}}{y^{3}}=\frac{\frac{c}{z}}{z^{3}} \\ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1 \end{cases}$$\Leftrightarrow\begin{cases}\frac{a}{x^4}=\frac{b}{y^4}=\frac{c}{z^4}
\\\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1 \end{cases}$
$\Leftrightarrow\begin{cases}\frac{x}{\sqrt[4]{a}}=\frac{y}{\sqrt[4]{b}}=\frac{z}{\sqrt[4]{c}}
\\ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1 \end{cases}$
$\Rightarrow\frac{a}{x}+\frac{b}{x\sqrt[4]{\frac{b}{a}}}+\frac{c}{x\sqrt[4]{\frac{c}{a}}}=1$ $\Rightarrow
x=\sqrt[4]{a}(\sqrt[4]{a^3}+\sqrt[4]{b^3}+\sqrt[4]{c^3})$
Tương tự :
$\begin{cases}y=\sqrt[4]{b}(\sqrt[4]{a^{3}}+\sqrt[4]{b^{3}}+\sqrt[4]{c^{3}}) \\z=\sqrt[4]{c}(\sqrt[4]{a^{3}}+\sqrt[4]{b^{3}}+\sqrt[4]{c^{3}})\end{cases}$
Vậy: $Min(S)=(\sqrt[4]{a^{3}}+\sqrt[4]{b^{3}}+\sqrt[4]{c^{3}})^{4}$