Optimization Contest Problem: Find the maximum of f(x): Why The Second Solution Fails?

Find the maximum of $f(x) = \dfrac{x^4-x^2}{x^6+2x^3-1}$ where $x>1$.

Previously, we discussed how to avoid using the surest and safest way of approaching this particular problem with the calculus method and opted for the inequality method.

To recap, here is what we have gotten:

$\begin{align*}f(x)&= \dfrac{x^4-x^2}{x^6+2x^3-1}\\&=\dfrac{1}{\dfrac{x^6+2x^3-1}{x^4-x^2}}\\&=\dfrac{1}{x^2+1+\dfrac{2x^3+x^2-1}{x^4-x^2}}\\&=\dfrac{1}{\dfrac{x^2-1}{2}+\dfrac{x^2-1}{2}+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}+2}\\&\le \dfrac{1}{5\left(\dfrac{x^2-1}{2}\cdot \dfrac{x^2-1}{2}\cdot \dfrac{1}{x^2}\cdot \dfrac{x}{x^2-1}\cdot \dfrac{x}{x^2-1}\right)^{\frac{1}{5}}+2}\\&\le \dfrac{1}{5\left(\dfrac{\cancel{x^2-1}}{2}\cdot \dfrac{\cancel{x^2-1}}{2}\cdot \dfrac{1}{\cancel{x^2}}\cdot \dfrac{\cancel{x}}{\cancel{x^2-1}}\cdot \dfrac{\cancel{x}}{\cancel{x^2-1}}\right)^{\frac{1}{5}}+2}\\&\le \dfrac{1}{5\left(\dfrac{1}{4}\right)^{\frac{1}{5}}+2}\end{align*}$

So the maximum of $f(x)$ that could be attained is $\dfrac{1}{5\left(\dfrac{1}{4}\right)^{\frac{1}{5}}+2}$. But we claimed that that maximum is not correct.

WHY! We cannot say it isn't correct and ditch it for good, as a mathematics proficient educator/student, you need to acquire the ability to tell and explain why the above is not a sound solution.

Note that we haven't checked the condition when does the equality occurs! Yeap! Yes! That is the point!

Equality occurs iff $\dfrac{x^2-1}{2}=\dfrac{1}{x^2}=\dfrac{x}{x^2-1}$.

In other words, no matter how we solve the above for $x$, the $x$ value must be the same...but, a quick check tells us that:

$\dfrac{x^2-1}{2}=\dfrac{1}{x^2}$

$x^4-x^2=2$

$x^4-x^2-2=0$

$(x^2-2)(x^2+1)=0\,\,\,\implies\,\,\,x=\sqrt{2}$ since $x\gt 1$.

If we replace $x=\sqrt{2}$ in $\dfrac{1}{x^2}=\dfrac{x}{x^2-1}$, we realize that

$\dfrac{1}{\sqrt{2}^2} \stackrel{?}=\dfrac{\sqrt{2}}{\sqrt{2}^2-1}$

$\dfrac{1}{2} \ne \sqrt{2}$

That means we have messed up somewhere while we were trying to regroup and rewrite the expression of $\dfrac{1}{x^2+1+\dfrac{2x^3+x^2-1}{x^4-x^2}}=\dfrac{1}{\dfrac{x^2-1}{2}+\dfrac{x^2-1}{2}+\dfrac{1}{x^2}+\dfrac{x}{x^2-1}+\dfrac{x}{x^2-1}+2}$.

We have to go back to the drawing board and start all over again. We would do the entire regrouping of the terms again, believe me, this looks like we have reached an impasse but we must believe in ourselves that with perseverance we can succeed. As trite and dull as this adage goes, it is also true.

I want to encourage you to try the problem out, as many times as long as you have not yet losing your patient before checking out the solution of mine that I am going to post it tomorrow.