Answer
(a) We show that the set of critical points of $f\left( {x,y} \right)$ is the curve $y = {x^{\alpha - 1}}$ (as given in Figure 28) or the curve $x = {y^{\beta - 1}}$.
The value of $f\left( {x,y} \right)$ at points on the curve $y = {x^{\alpha - 1}}$ is zero.
The value of $f\left( {x,y} \right)$ at points on the curve $x = {y^{\beta - 1}}$ is zero.
(b) Evaluating the discriminant, we show that the Second Derivative Test fails.
Using different approach we show that
$g\left( x \right) = f\left( {x,b} \right)$ is concave up with the critical point at $x = {b^{\beta - 1}}$. This implies that $f\left( {{b^{\beta - 1}},b} \right)$$f\left( {{b^{\beta - 1}},b} \right)$ is a local minimum.
(c) We show that $f\left( {{b^{\beta - 1}},b} \right) = 0$ is a minimum value, hence conclude that for all $x>0$, $f\left( {x,b} \right) \ge f\left( {{b^{\beta - 1}},b} \right) = 0$.
Using the results from part (a) and (b) we prove that
$\frac{1}{\alpha }{x^\alpha } + \frac{1}{\beta }{x^\beta } \ge xy$
Work Step by Step
(a) Consider the function $f\left( {x,y} \right) = {\alpha ^{ - 1}}{x^\alpha } + {\beta ^{ - 1}}{y^\beta } - xy$ that satisfies $f\left( {x,y} \right) \ge 0$ for all $x,y \ge 0$.
The partial derivatives are
${f_x} = {x^{\alpha - 1}} - y$, ${\ \ \ }$ ${f_y} = {y^{\beta - 1}} - x$
To find the critical points of $f$ we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = {x^{\alpha - 1}} - y = 0$, ${\ \ \ }$ ${f_y} = {y^{\beta - 1}} - x = 0$
The first and the second equation gives $y = {x^{\alpha - 1}}$ and $x = {y^{\beta - 1}}$, respectively.
Thus, the set of critical points of $f\left( {x,y} \right)$ is the curve $y = {x^{\alpha - 1}}$ (as given in Figure 28) or the curve $x = {y^{\beta - 1}}$.
From $y = {x^{\alpha - 1}}$ we obtain $x = {y^{1/\left( {\alpha - 1} \right)}}$.
Since we also have $x = {y^{\beta - 1}}$, therefore
$\beta - 1 = \frac{1}{{\alpha - 1}}$
(1) ${\ \ \ \ }$ $\left( {\beta - 1} \right)\left( {\alpha - 1} \right) = 1$
$\alpha \beta - \left( {\alpha + \beta } \right) + 1 = 1$
(2) ${\ \ \ \ }$ $\alpha \beta = \alpha + \beta $
$1 = \frac{{\alpha + \beta }}{{\alpha \beta }} = \frac{1}{\beta } + \frac{1}{\alpha }$
Hence, ${\alpha ^{ - 1}} + {\beta ^{ - 1}} = 1$.
The value of $f\left( {x,y} \right)$ at the curve $y = {x^{\alpha - 1}}$ is
$f\left( {x,{x^{\alpha - 1}}} \right) = {\alpha ^{ - 1}}{x^\alpha } + {\beta ^{ - 1}}{\left( {{x^{\alpha - 1}}} \right)^\beta } - x\left( {{x^{\alpha - 1}}} \right)$
$ = {\alpha ^{ - 1}}{x^\alpha } + {\beta ^{ - 1}}{x^{\alpha \beta - \beta }} - {x^\alpha }$
From equation (2) we get $\alpha \beta - \beta = \alpha $. Thus,
$f\left( {x,{x^{\alpha - 1}}} \right) = {\alpha ^{ - 1}}{x^\alpha } + {\beta ^{ - 1}}{x^\alpha } - {x^\alpha }$
$ = {x^\alpha }\left( {{\alpha ^{ - 1}} + {\beta ^{ - 1}} - 1} \right)$
But ${\alpha ^{ - 1}} + {\beta ^{ - 1}} = 1$. Therefore, $f\left( {x,{x^{\alpha - 1}}} \right) = 0$.
Thus, the value of $f\left( {x,y} \right)$ at points on the curve $y = {x^{\alpha - 1}}$ is zero.
Similarly,
$f\left( {{y^{\beta - 1}},y} \right) = {\alpha ^{ - 1}}{\left( {{y^{\beta - 1}}} \right)^\alpha } + {\beta ^{ - 1}}{y^\beta } - \left( {{y^{\beta - 1}}} \right)y$
$ = {\alpha ^{ - 1}}{y^{\alpha \beta - \alpha }} + {\beta ^{ - 1}}{y^\beta } - {y^\beta }$
From equation (2) we get $\alpha \beta - \alpha = \beta $. Thus,
$f\left( {{y^{\beta - 1}},y} \right) = {\alpha ^{ - 1}}{y^\beta } + {\beta ^{ - 1}}{y^\beta } - {y^\beta }$
$ = {y^\beta }\left( {{\alpha ^{ - 1}} + {\beta ^{ - 1}} - 1} \right)$
Using ${\alpha ^{ - 1}} + {\beta ^{ - 1}} = 1$, we obtain $f\left( {{y^{\beta - 1}},y} \right) = 0$.
Thus, the value of $f\left( {x,y} \right)$ at points on the curve $x = {y^{\beta - 1}}$ is zero.
(b) The second partial derivatives of $f$ are
${f_{xx}} = \left( {\alpha - 1} \right){x^{\alpha - 2}}$, ${\ \ }$ ${f_{yy}} = \left( {\beta - 1} \right){y^{\beta - 2}}$, ${\ \ }$ ${f_{xy}} = - 1$
The discriminant is
$D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = \left( {\alpha - 1} \right)\left( {\beta - 1} \right){x^{\alpha - 2}}{y^{\beta - 2}} - 1$
At the critical points $\left( {x,{x^{\alpha - 1}}} \right)$:
$D = \left( {\alpha - 1} \right)\left( {\beta - 1} \right){x^{\alpha - 2}}{\left( {{x^{\alpha - 1}}} \right)^{\beta - 2}} - 1$
$ = \left( {\alpha - 1} \right)\left( {\beta - 1} \right){x^{\alpha - 2}}{\left( {{x^{\alpha - 1}}} \right)^{\beta - 1}}{\left( {{x^{\alpha - 1}}} \right)^{ - 1}} - 1$
$ = \left( {\alpha - 1} \right)\left( {\beta - 1} \right){x^{\alpha - 2}}{x^{\left( {\alpha - 1} \right)\left( {\beta - 1} \right)}}{\left( {{x^{\alpha - 1}}} \right)^{ - 1}} - 1$
From equation (1) in part (a) we get $\left( {\beta - 1} \right)\left( {\alpha - 1} \right) = 1$. Thus,
$D = {x^{\alpha - 2}}x{\left( {{x^{\alpha - 1}}} \right)^{ - 1}} - 1 = {x^{\alpha - 1}}{\left( {{x^{\alpha - 1}}} \right)^{ - 1}} - 1 = 0$
Since $D=0$, by Theorem 2, the Second Derivative Test fails.
To find the extremum, we try a different approach:
For fixed $b>0$, we can write
$g\left( x \right) = f\left( {x,b} \right) = {\alpha ^{ - 1}}{x^\alpha } + {\beta ^{ - 1}}{b^\beta } - bx$
The derivatives are
$g'\left( x \right) = {x^{\alpha - 1}} - b$, ${\ \ }$ $g{\rm{''}}\left( x \right) = \left( {\alpha - 1} \right){x^{\alpha - 2}}$
To find the critical points of $g$ we solve the equation $g'\left( x \right) = 0$:
$g'\left( x \right) = {x^{\alpha - 1}} - b = 0$
$b = {x^{\alpha - 1}}$
$x = {b^{1/\left( {\alpha - 1} \right)}}$
Using equation (1) in part (a) we get $\beta - 1 = \frac{1}{{\alpha - 1}}$. Thus, the critical point of $g$ is at $x = {b^{\beta - 1}}$.
Since $\alpha \ge 1$, so $g{\rm{''}}\left( {{b^{\beta - 1}}} \right) = \left( {\alpha - 1} \right){\left( {{b^{\beta - 1}}} \right)^{\alpha - 2}} \ge 0$.
Hence, the function $g\left( x \right) = f\left( {x,b} \right)$ is concave up with the critical point at $x = {b^{\beta - 1}}$. This implies that $f\left( {{b^{\beta - 1}},b} \right)$ is a local minimum.
(c) From part (b) we know that $f\left( {{b^{\beta - 1}},b} \right)$ is a local minimum.
Recall from part (a), we have shown that $f\left( {{y^{\beta - 1}},y} \right) = 0$. Hence, $f\left( {{b^{\beta - 1}},b} \right) = 0$ is the minimum value. Since $f\left( {{b^{\beta - 1}},b} \right)$ is a local minimum, we conclude that for all $x>0$, $f\left( {x,b} \right) \ge f\left( {{b^{\beta - 1}},b} \right) = 0$.
Since $b>0$ is arbitrary, it follows that $f\left( {x,y} \right) \ge 0$. So,
$f\left( {x,y} \right) = {\alpha ^{ - 1}}{x^\alpha } + {\beta ^{ - 1}}{y^\beta } - xy \ge 0$
Hence, $\frac{1}{\alpha }{x^\alpha } + \frac{1}{\beta }{x^\beta } \ge xy$.
