Answer
First, we show that the maximum of $f\left( {{x_1},...,{x_n}} \right) = {x_1}{x_2}\cdot\cdot\cdot{x_n}$ subject to the constraints occurs for ${x_1} = {x_2} = \cdot\cdot\cdot = {x_n} = B/n$.
We use this result to conclude that:
${\left( {{a_1}{a_2}\cdot\cdot\cdot{a_n}} \right)^{1/n}} \le \frac{{{a_1} + {a_2} + \cdot\cdot\cdot + {a_n}}}{n}$
for all positive numbers ${a_1},...,{a_n}$.
Work Step by Step
We are given a function $f\left( {{x_1},...,{x_n}} \right) = {x_1}{x_2}\cdot\cdot\cdot{x_n}$ subject to the constraints $g\left( {{x_1},...,{x_n}} \right) = {x_1} + {x_2} + \cdot\cdot\cdot + {x_n} - B = 0$ and ${x_j} \ge 0$ for $j = 1,...,n$.
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
(1) ${\ \ \ }$ $\left( {{x_2}{x_3}\cdot\cdot\cdot{x_n},{x_1}{x_3}\cdot\cdot\cdot{x_n},...,{x_1}{x_2}\cdot\cdot\cdot{x_{n - 1}}} \right) = \lambda \left( {1,1,...,1} \right)$
So, the Lagrange equations are
$\begin{array}{*{20}{c}}
{{x_2}{x_3}\cdot\cdot\cdot{x_n} = \lambda }\\
{{x_1}{x_3}\cdot\cdot\cdot{x_n} = \lambda }\\
{...}\\
{{x_1}{x_2}\cdot\cdot\cdot{x_{n - 1}} = \lambda }
\end{array}$
if $\lambda = 0$, then equation (1) implies that ${x_j} = 0$ for $j = 1,...,n$. Since $\left( {0,0,...,0} \right)$ does not satisfy the constraint, we may assume that ${x_j} > 0$ for $j = 1,...,n$. So,
$\lambda = {x_2}{x_3}\cdot\cdot\cdot{x_n} = {x_1}{x_3}\cdot\cdot\cdot{x_n} = \cdot\cdot\cdot = {x_1}{x_2}\cdot\cdot\cdot{x_{n - 1}}$
We may write
$\begin{array}{*{20}{c}}
{{x_2}{x_3}\cdot\cdot\cdot{x_n} = {x_1}{x_3}\cdot\cdot\cdot{x_n}}\\
{{x_1}{x_3}\cdot\cdot\cdot{x_n} = {x_1}{x_2}\cdot\cdot\cdot{x_n}}\\
{\cdot\cdot\cdot}\\
{{x_1}{x_2}\cdot\cdot\cdot{x_n} = {x_1}{x_2}\cdot\cdot\cdot{x_{n - 1}}}
\end{array}$
Dividing common terms on both sides we get
$\begin{array}{*{20}{c}}
{{x_2} = {x_1}}\\
{{x_3} = {x_2}}\\
{...}\\
{{x_n} = {x_{n - 1}}}
\end{array}$
Thus, ${x_1} = {x_2} = \cdot\cdot\cdot = {x_n}$.
Using the constraints ${x_1} + {x_2} + \cdot\cdot\cdot + {x_n} = B$, we obtain the solution
$n{x_1} = n{x_2} = \cdot\cdot\cdot = n{x_n} = B$
${x_1} = {x_2} = \cdot\cdot\cdot = {x_n} = B/n$
So, the critical point is $\left( {{x_1},{x_2},...,{x_n}} \right) = \left( {\frac{B}{n},\frac{B}{n},...,\frac{B}{n}} \right)$.
The extreme value corresponding to $\left( {\frac{B}{n},\frac{B}{n},...,\frac{B}{n}} \right)$ is
$f\left( {\frac{B}{n},\frac{B}{n},...,\frac{B}{n}} \right) = {\left( {\frac{B}{n}} \right)^n}$
Since ${x_1},{x_2},...,{x_n} > 0$, the function $f\left( {{x_1},...,{x_n}} \right) = {x_1}{x_2}\cdot\cdot\cdot{x_n}$ is increasing. So, we conclude that the maximum of $f\left( {{x_1},...,{x_n}} \right)$ subject to the constraint is equal to ${\left( {\frac{B}{n}} \right)^n}$.
Hence, the maximum of $f\left( {{x_1},...,{x_n}} \right) = {x_1}{x_2}\cdot\cdot\cdot{x_n}$ subject to the constraints ${x_1} + {x_2} + \cdot\cdot\cdot + {x_n} = B$ and ${x_j} \ge 0$ for $j = 1,...,n$ occurs for ${x_1} = {x_2} = \cdot\cdot\cdot = {x_n} = B/n$.
Next, we use this result to conclude that
${\left( {{a_1}{a_2}\cdot\cdot\cdot{a_n}} \right)^{1/n}} \le \frac{{{a_1} + {a_2} + \cdot\cdot\cdot + {a_n}}}{n}$
for all positive numbers ${a_1},...,{a_n}$.
Consider some positive numbers ${a_1},...,{a_n}$. Since the maximum of $f\left( {{x_1},...,{x_n}} \right) = {x_1}{x_2}\cdot\cdot\cdot{x_n}$ is ${\left( {\frac{B}{n}} \right)^n}$, then
${a_1}{a_2}\cdot\cdot\cdot{a_n} \le {\left( {\frac{B}{n}} \right)^n}$
Since ${a_1} + {a_2} + \cdot\cdot\cdot + {a_n} = B$, so
${a_1}{a_2}\cdot\cdot\cdot{a_n} \le {\left( {\frac{{{a_1} + {a_2} + \cdot\cdot\cdot + {a_n}}}{n}} \right)^n}$
Hence, ${\left( {{a_1}{a_2}\cdot\cdot\cdot{a_n}} \right)^{1/n}} \le \frac{{{a_1} + {a_2} + \cdot\cdot\cdot + {a_n}}}{n}$.
Since ${a_1},...,{a_n}$ are arbitrary, it applies to all positive numbers ${a_1},...,{a_n}$.
