Minimum value of the expression

Question

$$ \frac {(x + \frac1x)^6 - (x^6 + \frac1{x^6}) - 2} {(x + \frac1x)^3 - (x^3 + \frac1{x^3})} $$

I could think of things like AM $\geq$ GM but couldn't see how it would help me to get a maxima. All help would be appreciated. By the way, for some reason I am getting the correct answer when I substitute for each bracket using AM $\geq$ GM a value of 2.

Answer 1

We will assume $x >0$. Notice $$\begin{align} f(x) \stackrel{def}{=} &\; \frac{(x+\frac1x)^6 - (x^6 + \frac{1}{x^6}) - 2}{(x+\frac1x)^3 - (x^3 + \frac{1}{x^3})} = \frac{(x+\frac1x)^6 - (x^3 + \frac{1}{x^3})^2}{(x+\frac1x)^3 - (x^3 + \frac{1}{x^3})}\\ = &\;\left(x+\frac1x\right)^3 + \left(x^3 + \frac{1}{x^3}\right) \end{align} $$ Apply AM $\ge$ GM to the two backets, we have

$$\left(x+\frac1x\right)^3 \ge \left(2\sqrt{x \cdot \frac{1}{x}}\right)^3 = 2^3 = 8 \quad\text{ and }\quad \left(x^3 + \frac{1}{x^3}\right) \ge 2\sqrt{x^3\cdot \frac{1}{x^3}} = 2 $$ This leads to

$$f(x) \ge 8 + 2 = 10\quad\text{ for } x > 0$$

Since this value $10$ is achieved at $x = 1$, we obtain $$\min_{x > 0} f(x) = 10$$

Answer 2

Hint: simplifying the numerator we get $$3\,{\frac { \left( 2\,{x}^{4}+{x}^{2}+2 \right) \left( {x}^{2}+1 \right) ^{2}}{{x}^{4}}} $$ and the denominator $$3\,{\frac {{x}^{2}+1}{x}}$$

Answer 3

Let $\displaystyle y=x+\frac{1}{x}$.

\begin{align*} \frac {(x + \frac1x)^6 - (x^6 + \frac1{x^6}) - 2} {(x + \frac1x)^3 - (x^3 + \frac1{x^3})}& = \frac {(x + \frac1x)^6 - (x^3 + \frac1{x^3})^2 } {(x + \frac1x)^3 - (x^3 + \frac1{x^3})} \\ &=\left(x + \frac1x\right)^3 + \left(x^3 + \frac1{x^3}\right)\\ &=\left(x + \frac1x\right)^3 + \left(x + \frac1x\right)^3-3\left(x + \frac1x\right)\\ &=y(2y^2-3) \end{align*}

Note that $\displaystyle y\ge2\sqrt{x\left(\frac{1}{x}\right)}=2$ and $y(2y^2-3)$ is strictly increasing for $y\ge 2$.

The minimum value is $(2)[2(2)^2-3]=10$.