Answer
The maximum value of $f\left( {x,y,z} \right) = {x^a}{y^b}{z^c}$ is $\sqrt {\frac{{{a^a}{b^b}{c^c}}}{{{{\left( {a + b + c} \right)}^{a + b + c}}}}} $ on the unit sphere ${x^2} + {y^2} + {z^2} = 1$.
Work Step by Step
We are given $f\left( {x,y,z} \right) = {x^a}{y^b}{z^c}$ for $x,y,z \ge 0$, where $a,b,c > 0$ are constants.
Our task is to find the maximum value of $f$ subject to the constraint $g\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {a{x^{a - 1}}{y^b}{z^c},b{x^a}{y^{b - 1}}{z^c},c{x^a}{y^b}{z^{c - 1}}} \right) = \lambda \left( {2x,2y,2z} \right)$
So, the Lagrange equations are
(1) ${\ \ \ }$ $a{x^{a - 1}}{y^b}{z^c} = 2\lambda x$, ${\ \ }$ $b{x^a}{y^{b - 1}}{z^c} = 2\lambda y$, ${\ \ }$ $c{x^a}{y^b}{z^{c - 1}} = 2\lambda z$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
We have $x,y,z \ge 0$. Since $\left( {0,0,0} \right)$ does not satisfy the constraint, we have the following cases:
Case 1. $x=0$, $y=0$, $z \ne 0$
Substituting $x=0$, $y=0$ in the constraint gives $z = \pm 1$. Since $z \ge 0$, the critical point is $\left( {0,0,1} \right)$.
Case 2. $x \ne 0$, $y=0$, $z=0$
Substituting $y=0$, $z=0$ in the constraint gives $x = \pm 1$. Since $x \ge 0$, the critical point is $\left( {1,0,0} \right)$.
Case 3. $x=0$, $y \ne 0$, $z=0$
Substituting $x=0$, $z=0$ in the constraint gives $y = \pm 1$. Since $y \ge 0$, the critical point is $\left( {0,1,0} \right)$.
Case 4. $x \ne 0$, $y \ne 0$, $z \ne 0$
In this case, equation (1) implies that $\lambda \ne 0$. So,
$\lambda = \frac{a}{2}{x^{a - 2}}{y^b}{z^c} = \frac{b}{2}{x^a}{y^{b - 2}}{z^c} = \frac{c}{2}{x^a}{y^b}{z^{c - 2}}$
$\frac{{{y^b}}}{{{y^{b - 2}}}} = \frac{b}{a}\frac{{{x^a}}}{{{x^{a - 2}}}}$, ${\ \ }$ $\frac{{{z^c}}}{{{z^{c - 2}}}} = \frac{c}{b}\frac{{{y^b}}}{{{y^{b - 2}}}}$, ${\ \ }$ $\frac{{{x^a}}}{{{x^{a - 2}}}} = \frac{a}{c}\frac{{{z^c}}}{{{z^{c - 2}}}}$
${y^2} = \frac{b}{a}{x^2}$, ${\ \ \ }$ ${z^2} = \frac{c}{b}{y^2}$, ${\ \ \ }$ ${x^2} = \frac{a}{c}{z^2}$
So, $y = \pm \sqrt {\frac{b}{a}} x$, $z = \pm \sqrt {\frac{c}{b}} y$.
Thus, $z = \pm \sqrt {\frac{c}{a}} x$.
Step 3. Solve for $x$, $y$ and $z$ using the constraint
In Step 2, we obtain $y = \pm \sqrt {\frac{b}{a}} x$ and $z = \pm \sqrt {\frac{c}{a}} x$. Substituting them in the constraint gives
${x^2} + \frac{b}{a}{x^2} + \frac{c}{a}{x^2} - 1 = 0$
${x^2}\left( {\frac{{a + b + c}}{a}} \right) = 1$
So, $x = \pm \sqrt {\frac{a}{{a + b + c}}} $.
Using $y = \pm \sqrt {\frac{b}{a}} x$ and $z = \pm \sqrt {\frac{c}{a}} x$, we obtain
$y = \pm \sqrt {\frac{b}{{a + b + c}}} $ and $z = \pm \sqrt {\frac{c}{{a + b + c}}} $.
So, the solutions are $\left( { \pm \sqrt {\frac{a}{{a + b + c}}} , \pm \sqrt {\frac{b}{{a + b + c}}} , \pm \sqrt {\frac{c}{{a + b + c}}} } \right)$.
However, since $x,y,z \ge 0$ and $a,b,c > 0$, there is only one critical point at $\left( {\sqrt {\frac{a}{{a + b + c}}} ,\sqrt {\frac{b}{{a + b + c}}} ,\sqrt {\frac{c}{{a + b + c}}} } \right)$.
Step 4. Calculate the critical values
We evaluate the extreme values at the critical points and list them in the following table:
$\begin{array}{*{20}{c}}
{{\rm{Critical{\ }point}}}&{f\left( {x,y,z} \right)}\\
{\left( {0,0,1} \right)}&0\\
{\left( {1,0,0} \right)}&0\\
{\left( {0,1,0} \right)}&0\\
{\left( {\sqrt {\frac{a}{{a + b + c}}} ,\sqrt {\frac{b}{{a + b + c}}} ,\sqrt {\frac{c}{{a + b + c}}} } \right)}&{{{\left( {\sqrt {\frac{a}{{a + b + c}}} } \right)}^a}{{\left( {\sqrt {\frac{b}{{a + b + c}}} } \right)}^b}{{\left( {\sqrt {\frac{c}{{a + b + c}}} } \right)}^c}}
\end{array}$
From the results in this table, we conclude that the maximum value of $f$ is $\sqrt {\frac{{{a^a}{b^b}{c^c}}}{{{{\left( {a + b + c} \right)}^{a + b + c}}}}} $ on the unit sphere ${x^2} + {y^2} + {z^2} = 1$.
