Đặt: $a=a_{1}+a_{2}+...+a_{n}, b_{i}=\frac{a_{i}}{a}(i=1,2,...,n)$ thì $b_{i}>0$ và:
$b_{1}+b_{2}+...+b_{n}=1$
Áp dụng BĐT Cauchy suy rộng :
$(\frac{x_{1}}{a_{1}})^{b_{1}}.(\frac{x_{2}}{a_{2}})^{b_{2}}...(\frac{x_{n}}{a_{n}})^{b_{n}} \leq \frac{b_{1}}{a_{1}}.x_{1} +\frac{b_{2}}{a_{2}}.x_{2}+...+\frac{b_{n}}{a_{n}}.x_{n} =$
$= \frac{1}{a}(x_{1}+x_{2}+...+x_{n})= \frac{1}{a}$
$\Rightarrow x_1^{b_1}.x_2^{b_2}…x_n^{b_n}\leq\frac{1}{a}.a_1^{b_1}.a_2^{b_2}…a_n^{b_n}$
$\Leftrightarrow S= x_1^{a_1}.x_2^{a_2}…x_n^{a_n}\leq\frac{1}{a^a}.a_1^{a_1}.a_2^{a_2}…a_n^{a_n}$
Dấu "=" xảy ra $\Leftrightarrow$ $\frac{x_1}{a_1}=\frac{x_2}{a_2}=…=\frac{x_n}{a_n}=\frac{x_1+x_2+…+x_n}{a_1+a_2+…+a_n}=\frac{1}{a}$
( theo tính chất của dãy tỉ số bằng nhau)
$\Rightarrow x_i=\frac{a_i}{a}(\forall i=1, 2, …, n)$
Vậy: $Max(S)= \frac{1}{a^a}.a^{a_{1}}_{1}.a^{a_{2}}_{2}...a^{a_{n}}_{n}$