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 ${x_1}^2 + {x_2}^2 + \cdot\cdot\cdot + {x_n}^2 = {B^2}$ is $\sqrt n B$.
Using this result we conclude that
$\left| {{a_1}} \right| + \left| {{a_2}\left| { + \cdot\cdot\cdot + } \right|{a_n}} \right| \le \sqrt n {\left( {{a_1}^2 + {a_2}^2 + \cdot\cdot\cdot + {a_n}^2} \right)^{1/2}}$
for all numbers ${a_1},{a_2},...,{a_n}$.
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 $g\left( {{x_1},...,{x_n}} \right) = {x_1}^2 + {x_2}^2 + \cdot\cdot\cdot + {x_n}^2 - {B^2} = 0$.
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
$\begin{array}{*{20}{c}}
{{x_2} + {x_3} + \cdot\cdot\cdot + {x_n} = \lambda }\\
{{x_1} + {x_3}\cdot\cdot\cdot + {x_n} = \lambda }\\
{...}\\
{{x_1} + {x_2} + \cdot\cdot\cdot + {x_{n - 1}} = \lambda }
\end{array}$
Thus,
$\lambda = {x_2} + {x_3} + \cdot\cdot\cdot + {x_n} = {x_1} + {x_3}\cdot\cdot\cdot + {x_n} = ... = {x_1} + {x_2} + \cdot\cdot\cdot + {x_{n - 1}}$
We may write
$\begin{array}{*{20}{c}}
{{x_2} + {x_3} + \cdot\cdot\cdot + {x_n} = {x_1} + {x_3} + \cdot\cdot\cdot + {x_n}}\\
{{x_1} + {x_3} + \cdot\cdot\cdot + {x_n} = {x_1} + {x_2} + \cdot\cdot\cdot + {x_n}}\\
{\cdot\cdot\cdot}\\
{{x_1} + {x_2} + \cdot\cdot\cdot + {x_n} = {x_1} + {x_2} + \cdot\cdot\cdot + {x_{n - 1}}}
\end{array}$
Subtracting 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 ${x_1}^2 + {x_2}^2 + \cdot\cdot\cdot + {x_n}^2 = {B^2}$, we obtain the solution
$n{x_1}^2 = n{x_2}^2 = \cdot\cdot\cdot = n{x_n}^2 = {B^2}$
$\left| {{x_1}} \right| = \left| {{x_2}} \right| = \cdot\cdot\cdot = \frac{B}{{\sqrt n }}$
So, the critical point is $\left( {{x_1},{x_2},...,{x_n}} \right) = \left( {\frac{B}{{\sqrt n }},\frac{B}{{\sqrt n }},...,\frac{B}{{\sqrt n }}} \right)$.
The extreme value corresponding to $\left( {\frac{B}{{\sqrt n }},\frac{B}{{\sqrt n }},...,\frac{B}{{\sqrt n }}} \right)$ is
$f\left( {\frac{B}{{\sqrt n }},\frac{B}{{\sqrt n }},...,\frac{B}{{\sqrt n }}} \right) = n\frac{B}{{\sqrt n }} = \sqrt n B$
For positive numbers, ${x_1},{x_2},...,{x_n} > 0$, 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 budget constraint is equal to $\sqrt n B$.
Hence, the maximum of $f\left( {{x_1},...,{x_n}} \right) = {x_1} + {x_2}\cdot\cdot\cdot + {x_n}$ subject to ${x_1}^2 + {x_2}^2 + \cdot\cdot\cdot + {x_n}^2 = {B^2}$ is $\sqrt n B$.
Next, we use this result to conclude that
$\left| {{a_1}} \right| + \left| {{a_2}\left| { + \cdot\cdot\cdot + } \right|{a_n}} \right| \le \sqrt n {\left( {{a_1}^2 + {a_2}^2 + \cdot\cdot\cdot + {a_n}^2} \right)^{1/2}}$
for all numbers ${a_1},{a_2},...,{a_n}$.
Consider some numbers ${a_1},{a_2},...,{a_n}$. Since the maximum of $f\left( {{x_1},...,{x_n}} \right) = {x_1} + {x_2} + \cdot\cdot\cdot + {x_n}$ is $\sqrt n B$, then
$\left| {{a_1}} \right| + \left| {{a_2}\left| { + \cdot\cdot\cdot + } \right|{a_n}} \right| \le \sqrt n B$
Using ${a_1}^2 + {a_2}^2 + \cdot\cdot\cdot + {a_n}^2 = {B^2}$, we obtain
$\left| {{a_1}} \right| + \left| {{a_2}\left| { + \cdot\cdot\cdot + } \right|{a_n}} \right| \le \sqrt n {\left( {{a_1}^2 + {a_2}^2 + \cdot\cdot\cdot + {a_n}^2} \right)^{1/2}}$
Since ${a_1},{a_2},...,{a_n}$ are arbitrary, it applies to all numbers ${a_1},{a_2},...,{a_n}$.
