The minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$

Question

Problem :

The minimum value of $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$$

Can I use this in numerator and denominator :

The minimum value of $a +\frac{1}{a}$

Using A.M and G.M inequality :

$a +\frac{1}{a} \geq 2\sqrt{a \times \frac{1}{a}}$

$\Rightarrow a +\frac{1}{a} \geq 2$ .....(1)

By putting the minimum value of (1) in $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$

we get ; $\frac{2^6-2-2}{2^3+2}$ but I think this is wrong especially denominator as we need to find the maximum value of denominator to get the minimum value.

Please suggest ,thanks.

Answer 1

hint: let $$f(x)=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+x^{-3}}=3\left(x+\dfrac{1}{x}\right)\ge 6$$

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

Answer 2

Here is one way to go about it. We want to express$\left(x^6+\frac1{x^6}\right)$ in terms of $\left(x+\frac1x\right)$. $$\left(x+\frac1x\right)^6={x}^{6}+6{x}^{4}+15{x}^{2}+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac1{x^6}$$ We can remove the $x^4$ and $\frac1{x^4}$ terms by subtracting $6\left( x+\frac1x\right)^4$, which gives $$\left(x+\frac1x\right)^6-6\left( x+\frac1x\right)^4={x}^{6}-9{x}^{2}-16-\frac{9}{x^2}+\frac1{x^6}$$ Continuing this process and using the substitution $u=x+\frac1x$ leads to $$x^6+\frac1{x^6}=\left(x+\frac1x\right)^6-6\left(x+\frac1x\right)^4+9\left(x+\frac1x\right)^2-2=u^6-6u^4+9u^2-2$$ and similarly $$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)=u^3-3u$$

Now you have the function $$f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\frac{u^6-(u^6-6u^4+9u^2-2)-2}{u^3+(u^3-3u)}$$ $$=\frac{6u^4-9u^2}{2u^3-3u}=3u=3\left(x+\frac1x\right)$$

Since $x+\frac1x\geq2$, the minimum value of $f(x)$ is thus $6$.