Optimization problem on function

[juantheron]

Maximum value of expression $\displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1}\;,$ where $x>1$

 

[anemone]

My solution:

Spoiler

If we want to maximize $f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\left(\dfrac{x^6+2x^3-1}{x^4-x^2}\right)}$, this could be done if we are to find the minimum value for the expression $\dfrac{x^6+2x^3-1}{x^4-x^2}$.

Note that

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

We then apply the AM-GM to the first three terms of the expression (*) and that gives

$\left(x-\dfrac{1}{x}\right)^2+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}\ge 3$

(*Equality occurs when $x^2-x-1=\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0$.)

Therefore, the minimum value for $\dfrac{x^6+2x^3-1}{x^4-x^2}$ is $3+3=6$ and that results in a maximum of

$f(x)=\dfrac{x^4-x^2}{x^6+2x^3-1}=\dfrac{1}{\left(\dfrac{x^6+2x^3-1}{x^4-x^2}\right)}$ as $\dfrac{1}{6}$

.

 

[anemone]

Hi jacks again, :)

I want to apologize for not formulating my solution above in a very good way.:(

I forgot to mention that the term $\dfrac{1}{\left(x-\dfrac{1}{x}\right)}$ is a positive in the given domain where $x>1$, therefore, I could then apply the AM-GM on those terms to look for the minimum value for $\left(x-\dfrac{1}{x}\right)^2+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}+\dfrac{1}{\left(x-\dfrac{1}{x}\right)}$.

Sorry about that.

 

[juantheron]

Thanks anemone for Nice Solution::

My Solution is Similar To Yours::

Spoiler

Given $$\displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1}\;,$$ Where $$x>1$$

So We can Simplify $$\displaystyle f(x) = \frac{x^4-x^2}{x^6+2x^3-1} = \frac{x^3\cdot \left(x-\frac{1}{x}\right)}{x^3\cdot \left(x^3-\frac{1}{x^3}\right)+2} = \frac{\left(x-\frac{1}{x}\right)}{\left(x^3-\frac{1}{x^3}\right)+2}$$

Now Let $$u = \left(x-\frac{1}{x}\right)>0\;,\left(x^3-\frac{1}{x^3}\right) = \left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)\;,x>1.$$

So $$\displaystyle f(u) = \frac{u}{u^3+3u+2}=\frac{u}{u^3+u+u+u+1+1}\leq \frac{u}{6\sqrt[6]{u^3\cdot u \cdot \cdot u \cdot u\cdot 1\cdot 1}} = \frac{1}{6}$$

Using $$\bf{A.M\geq G.M}$$ and above equality hold when $$\displaystyle u = 1\Rightarrow \left(x-\frac{1}{x}\right) = 1\Rightarrow x= \frac{\sqrt{5}+1}{2}$$