Minimum without computing the derivative.

Question

I am trying to find the minimum of the following function: $$H(x)=\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})},\hspace{1cm}x>0$$

What I usually do to find extrema of a function is computing the derivative of the function and finding its zeros. However, I think there must be a more efficient approach to this problem, since the derivative of this function is quite long. Any tips?

Thanks.

Answer 1

$$\displaystyle \bigg[\bigg(x+\frac{1}{x}\bigg)^3\bigg]^2-\bigg[x^3+\frac{1}{x^3}\bigg]^2$$

$$\bigg[\bigg(x+\frac{1}{x}\bigg)^3+x^3+\frac{1}{x^3}\bigg]\bigg[\bigg(x+\frac{1}{x}\bigg)^3-x^3-\frac{1}{x^3}\bigg]$$

$$H(x)=\bigg(x+\frac{1}{x}\bigg)^3-x^3-\frac{1}{x^3}=3\bigg(x+\frac{1}{x}\bigg)\geq 3\cdot 2=6$$

Because arithmetic geometric inequality

$$x+\frac{1}{x}\geq 2\;\forall \;x>0$$

Answer 2

Observe that $$H(x) = \frac{\left(\left(x+\frac{1}{x}\right)^3\right)^2 - \left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3 + \left(x^3+\frac{1}{x^3}\right)} = \left(x+\frac{1}{x}\right)^3 - \left(x^3+\frac{1}{x^3}\right) = 3\left(x+\frac{1}{x} \right) \geq 6$$ (with equality at $x=1$) where in the last step I have use the inequalities $x>0$ and $(x-1)^2 \geq 0$. On the other hand since $lim_{x \rightarrow 0} H(x) = \infty$, the range of $H$ is $[6, \infty)$.

Edit: Typos corrected, thanks to Rob for pointing them out, apologies for the delay.