# Determine the Minimum Value

#### anemone

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

#### Ackbach

##### Indicium Physicus
Staff member
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.

#### Jester

##### Well-known member
MHB Math Helper
Here's my solution

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.

#### Jester

##### Well-known member
MHB Math Helper
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
If we write your function as

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

It becomes obvious that the minimum is 8

#### anemone

##### MHB POTW Director
Staff member
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...

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

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

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

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

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

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

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

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