MHB What is the AM-GM Inequality Used For?

[anemone]

Determine the minimum value of $$\frac{\sec^4 a}{\tan^2 b}+\frac{\sec^4 b}{\tan^2 a}$$ over all $a,b \ne \frac{k \pi}{2}$ where $k$ is in $Z$.

 

[Ackbach]

Spoiler

Since $a$ and $b$ are symmetric in the expression, I will set $a=b$, and then define
$$f(a)= \frac{2 \sec^{4}(a)}{ \tan^{2}(a)}= \frac{2}{ \cos^{2}(a) \sin^{2}(a)}.$$
Differentiating yields
\begin{align*}
f'(a)&= \frac{-2(-2 \cos(a) \sin^{3}(a)+2 \sin(a) \cos^{3}(a))}{ \cos^{4}(a) \sin^{4}(a)} \\
&= \frac{-4 \sin(a) \cos(a) [ \cos^{2}(a)- \sin^{2}(a)]}{ \sin^{4}(a) \cos^{4}(a)} \\
&= \frac{-4 \cos(2a)}{ \sin^{3}(a) \cos^{3}(a)}.
\end{align*}
Setting $f'(a)=0$ implies that
$$2a= \frac{(2j+1) \pi}{2} \implies a=\frac{(2j+1) \pi}{4}.$$
Just to make sure, let us look for values of $a\in (0, \pi) \setminus \{\pi/2 \}$. This turns out to require $j=0,1$, and therefore $a\in \{ \pi/4, 3 \pi/4 \}$. Plugging either of these into $f$ yields the minimum value of $8$.

To be complete, we should take the second derivative $f''(a)$, and show that it is positive at these values. I will leave that to the reader.

 

[Mathwebster]

Here's my solution

Spoiler

Using the usual identity we have

$\dfrac{(\tan^2 a + 1)^2}{\tan^2b}+\dfrac{(\tan^2 b + 1)^2}{\tan^2a}.$

If we let $x = \tan a$ and $y = \tan b$ then we have

$z = \dfrac{(x^2 + 1)^2}{y^2}+\dfrac{(y^2 + 1)^2}{x}.$

Using the standard first derivatives we have (noting that $x,y \ne 0$)

$\dfrac{\partial z}{\partial x} = 4\,{\dfrac { \left( {x}^{2}+1 \right) x}{{y}^{2}}}-2\,{\dfrac { \left( {
y}^{2}+1 \right) ^{2}}{{x}^{3}}}$

$\dfrac{\partial z}{\partial y} = 4\,{\dfrac { \left( {y}^{2}+1 \right) y}{{x}^{2}}}-2\,{\dfrac { \left( {
x}^{2}+1 \right) ^{2}}{{y}^{3}}}$

Simplify and setting these to zero gives

$
\begin{align}
2\,{x}^{6}+2\,{x}^{4}-{y}^{6}-2\,{y}^{4}-{y}^{2} &= 0\;\;\;(*)\\
-{x}^{6}-2\,{x}^{4}-{x}^{2}+2\,{y}^{6}+2\,{y}^{4}&=0
\end{align}$

Multiplying the first by $x^2+1$ and the second by $2x^2$ and adding gives

${y}^{2} \left( {y}^{2}+1 \right) \left( 3\,{y}^{2}{x}^{2}-{y}^{2}-1-{
x}^{2} \right) =0$

from which we can solve for $y^2$ giving $y^2 = \dfrac{x^2+1}{3x^2-1}$ noting that $3x^2-1 \ne 0$. Substituting into (*) and factoring gives $x = \pm 1$ which in turn gives $y = \pm 1$ giving the minimum value of $z$ as $8$. The second derivative test verifies this.

 

[Mathwebster]

Spoiler

Since $a$ and $b$ are symmetric in the expression, I will set $a=b$, and then define
$$f(a)= \frac{2 \sec^{4}(a)}{ \tan^{2}(a)}= \frac{2}{ \cos^{2}(a) \sin^{2}(a)}.$$
Differentiating yields
\begin{align*}
f'(a)&= \frac{-2(-2 \cos(a) \sin^{3}(a)+2 \sin(a) \cos^{3}(a))}{ \cos^{4}(a) \sin^{4}(a)} \\
&= \frac{-4 \sin(a) \cos(a) [ \cos^{2}(a)- \sin^{2}(a)]}{ \sin^{4}(a) \cos^{4}(a)} \\
&= \frac{-4 \cos(2a)}{ \sin^{3}(a) \cos^{3}(a)}.
\end{align*}
Setting $f'(a)=0$ implies that
$$2a= \frac{(2j+1) \pi}{2} \implies a=\frac{(2j+1) \pi}{4}.$$
Just to make sure, let us look for values of $a\in (0, \pi) \setminus \{\pi/2 \}$. This turns out to require $j=0,1$, and therefore $a\in \{ \pi/4, 3 \pi/4 \}$. Plugging either of these into $f$ yields the minimum value of $8$.

To be complete, we should take the second derivative $f''(a)$, and show that it is positive at these values. I will leave that to the reader.

Just a note

Spoiler

If we write your function as

$f(a) = 8 \csc^2 2a$

It becomes obvious that the minimum is 8

 

[anemone]

Hi Ackbach and Jester,

Thank you for participating...and yes, the answer is correct.

The solution that comes along with this particular problem suggests the use of the AM-GM inequality to solve it...let's see...

Spoiler

If we let $x = \tan^2 a$ and $y = \tan^2 b$ then we have

$$\frac{(\tan^2 a + 1)^2}{\tan^2b}+\frac{(\tan^2 b + 1)^2}{\tan^2a} = \frac{(x + 1)^2}{y}+\frac{(y + 1)^2}{x}$$

$$\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{x^2 + 2x+1}{y}+\frac{y^2 +2y+ 1}{x}$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \left(\frac{x^2}{y}+\frac{1}{y}+\frac{x^2}{y}+ \frac{1}{x}\right)+2(\frac{x}{y}+\frac{y}{x})$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge 4\sqrt[4]{\frac{x^2}{y}\cdot\frac{1}{y}\cdot\frac{x^2}{y} \cdot\frac{1}{x}}+2\left(2\sqrt{\frac{x}{y}\cdot \frac{y}{x}}\right)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge 4+4$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ge 8$$

Equality holds when $x=y=1$, i.e. $a=\pm 45^{\circ}+k\cdot180^{\circ}$, $b=\pm 45^{\circ}+k\cdot180^{\circ}$ for integer $k$.