In my previous post(optimization-contest-problem), I showed you the way of approaching this particular problem with the most comfortable and familiar method (calculus method) that it would spring first to students' mind, a method that students' from all four corners of the world are inclined to use.
But I am fairly certain that you hope to look for another shortcut, or more elegant solution to avoid factoring the first derivative just to look for the critical points. Calculus way is the surest and safest way to obtain the valid answer for optimization problem, but, it has its downside too, take for example, it is not always feasible to obtain the critical points after we gotten the first derivative and set it to zero. If we fail to achieve that, then there is no way to continue with the calculus method to look for the optimize solution and we have to adopt other approaches.
Well, yes, this problem can be solved with the Calculus method, albeit the factoring process took like forever to finish.
That is why I am sorely tempted to lead you into even more alluring method and discovering, but that comes with endless trials (a gross exaggeration if you are not great at algebraically manipulating the mathematical expression), but worry no more, students, I will show you how to fix things from wrong to right and eventually to the desirable answer.
Let's not beat about the bush, let's get started!
The other method that we could adopt is the AM-GM inequality formula, AM-GM inequality tells us
1. If we have two positive real $a$ and $b$, then we too have
[MATH]\color{yellow}\bbox[5px,purple]{a+b\ge 2(ab)^{\frac{1}{2}}}[/MATH] with equality at $a=b$.
2. If we have three positive real $a$, $b$ and $c$ then we too have
[MATH]\color{yellow}\bbox[5px,green]{a+b+c\ge 3(ab)^{\frac{1}{3}}}[/MATH] with equality at $a=b=c$.
3. If we have two positive real $a$ and $\dfrac{1}{a}$, then we too have
[MATH]\color{black}\bbox[5px,orange]{a+\dfrac{1}{a}\ge 2\left(a\cdot \dfrac{1}{a}\right)^{\frac{1}{2}}=2}[/MATH] with equality at $a=\dfrac{1}{a}$.
This indeed is a very powerful result.
In other words, if we have $n$ positive real numbers, then AM-GM inequality says we have
$a_1+a_2+\cdots+a_n\ge n(a_1a_2\cdots a_n)^{\frac{1}{n}}$, with equality if $a_1=a_2=\cdots=a_n$.
Two things that we have to bear in mind when our method of attacking is all about the AM-GM inequality, one is we must ensure we are to deal with all the positive terms (in our case, the positive terms in the domain $x\gt 1$).
Second, we have to obtain the sum of various positive terms but if we have it the pairs of the positive terms with their respective reciprocals, then things would be simplified so neatly!
But what we have now is the rational function, [MATH]\color{yellow}\bbox[5px,blue]{ \dfrac{x^4-x^2}{x^6+2x^3-1}}[/MATH], how we are going to "break" it and rewrite it so we have a sum of some positive terms?
[MATH]\color{black}\bbox[5pxpink]{ \dfrac{x^4-x^2}{x^6+2x^3-1}}[/MATH] is a proper rational polynomial function, there is nothing much we can play with it, but, we can make it improper if we are to shift the expression on numerator down so we have:
[MATH]\color{black}\bbox[5px,pink]{ \dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}}[/MATH]
Note that we can do a lot to the improper rational polynomial function, the first and foremost that we could do is to perform the polynomial long division to get:
[MATH]\color{black}\bbox[5px,pink]{ \begin{align*}\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}&=\dfrac{1}{x^2+1+\dfrac{2x^3+x^2-1}{x^4-x^2}}\end{align*}}[/MATH]
Remember, things would be so much better if we could rewrite the expression so for each positive term we got, we would create a reciprocal of it...
Let's see what we could have thought of...
[MATH]\color{black}\bbox[5px,pink]{ \begin{align*}\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}&=\dfrac{1}{x^2+1+\dfrac{2x^3+x^2-1}{x^4-x^2}}\\&=\dfrac{1}{x^2+1+\dfrac{(2x^3+x^2)-1}{x^4-x^2}}\\&=\dfrac{1}{x^2+1+\dfrac{2x^3+x^2}{x^4-x^2}-\dfrac{1}{x^4-x^2}}\end{align*}}[/MATH]
We must stop now because we have grouped the wrong terms, we cannot afford to subtract anything if we want to use the AM-GM inequality but looking at the last term in the above equation, [MATH]\color{black}\bbox[5px,pink]{ -\dfrac{1}{x^4-x^2}}[/MATH] simply cannot exist.
So we have to try again by grouping differently than the first trial:
[MATH]\color{black}\bbox[5px,yellow]{ \begin{align*}\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}&=\dfrac{1}{x^2+1+\dfrac{2x^3+x^2-1}{x^4-x^2}}\\&=\dfrac{1}{x^2+1+\dfrac{2x^3+(x^2-1)}{x^4-x^2}}\\&=\dfrac{1}{x^2+1+\dfrac{2x^3}{x^2(x^2-1)}+\dfrac{x^2-1}{x^2(x^2-1)}}\\&=\dfrac{1}{x^2+1+\dfrac{2x}{x^2-1}+\dfrac{1}{x^2}}\end{align*}}[/MATH]
You might be happy for a moment as you can see it now that we have the reciprocal pair for $x^2$ and $\dfrac{1}{x^2}$. Little did you know that chances are bright too if those are not meant to be together...
[MATH]\color{black}\bbox[5px,yellow]{ \begin{align*}\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}&=\dfrac{1}{x^2+1+\dfrac{2x}{x^2-1}+\dfrac{1}{x^2}}\\&=\dfrac{1}{x^2+\dfrac{1}{x^2}+\dfrac{2x}{x^2-1}+1}\end{align*}}[/MATH]
We have hit a bumpy road now since there is no way to algebraically manipulating the remaining two terms $\dfrac{2x}{x^2-1}+1$ so when we applied the AM-GM formula we could cancel them out and get something meaningful...but some students want to try further before they calling it a failure, okay, that is not a problem at all. They are right, we should not give up easily once we hit the wall, let's see what would we get if we continue with the above:
[MATH]\color{black}\bbox[5px,yellow]{ \begin{align*}\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}&==\dfrac{1}{x^2+1+\dfrac{2x}{x^2-1}+\dfrac{1}{x^2}}\\&=\dfrac{1}{x^2+\dfrac{1}{x^2}+\dfrac{2x}{x^2-1}+1}\\&=\dfrac{1}{x^2+\dfrac{1}{x^2}+\dfrac{1}{x+1}+\dfrac{1}{x-1}+1}\\&=\dfrac{1}{x^2+\dfrac{1}{x^2}+\dfrac{1}{x+1}+\dfrac{1}{x-1}+(x+1)+(x-1)-(2x)+1}\end{align*}}[/MATH]
Can you see how I introduced the two terms $x+1$ and $x-1$ in the above so we have the reciprocals pairs but that simultaneously put in a negative term, and I have to repeat the fact that we cannot afford to have subtract something if we are to use the AM-GM formula. See it now that this configuration of the terms failed completely?
We have to try another grouping instead of succumbing to defeat and declare that calculus is the easiest way out.
Note that we still have another trick up our sleeves!
We could try to algebraically manipulating the expression in the following brilliant way:
$\begin{align*}x^2+\dfrac{1}{x^2}+\dfrac{2x}{x^2-1}+1&=x^2+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}+1\\&=x^2-1+1+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}+1\\&=\dfrac{x^2-1}{2}+\dfrac{x^2-1}{2}+1+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}+1\\&=\dfrac{x^2-1}{2}+\dfrac{x^2-1}{2}+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}+2\end{align*}$
Note that we can cancel out all the terms, leaving only numbers for the above when we apply AM-GM formula to it:
$\begin{align*}\dfrac{x^2-1}{2}+\dfrac{x^2-1}{2}+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}&\ge 5\left(\dfrac{x^2-1}{2}\cdot \dfrac{x^2-1}{2}\cdot \dfrac{1}{x^2}\cdot \dfrac{x}{x^2-1}\cdot \dfrac{x}{x^2-1}\right)^{\frac{1}{5}}\\&\ge 5\left(\dfrac{\cancel{x^2-1}}{2}\cdot \dfrac{\cancel{x^2-1}}{2}\cdot \dfrac{1}{\cancel{x^2}}\cdot \dfrac{\cancel{x}}{\cancel{x^2-1}}\cdot \dfrac{\cancel{x}}{\cancel{x^2-1}}\right)^{\frac{1}{5}}\\&\ge 5\left(\dfrac{1}{4}\right)^{\frac{1}{5}}\end{align*}$
WOW! This seems so neat and pretty that it can't go wrong this one time, and you want to say it out loud that the minimum of [MATH]\color{black}\bbox[5pxpink]{ \dfrac{x^6+2x^3-1}{x^4-x^2}}[/MATH] in [MATH]\color{black}\bbox[5pxpink]{ f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}}[/MATH] is $5\left(\dfrac{1}{4}\right)^{\frac{1}{5}}$, that of course results in the maximum of $f(x)=\dfrac{1}{5\left(\dfrac{1}{4}\right)^{\frac{1}{5}}}$.
The thing is, do you think we are done by now?
I will let you think about it for a moment before telling you why this is a total failure and how there is always price to pay if we don't check and verify the final result.
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