Putnam 1998

$\begin{aligned} f(x)&=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\\ &=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}+2\right)}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\\ &=\dfrac{\left[\left(x+\dfrac{1}{x}\right)^3\right]^2-\left(x^3+\dfrac{1}{x^3}\right)^2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\\ &=\dfrac{\left[\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)\right]\cdot\left[\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)\right]}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\\ &=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)=3x+\dfrac{3}{x} \end{aligned}$

Taking the first derivative and setting it equal to $0$ to find the critical points

$f'(x)=3-\dfrac{3}{x^2}=0\iff x=\pm 1$

Since $x>0$ the critical point is at $x=1$

$f''(1)=\left.\dfrac{1}{x^3}\right|_{1}=1>0$

So this point is a minimum, hence

$\min\{f(x)\}=f(1)=\left. 3x+\dfrac{1}{x}\right|_{1} =\boxed{6}$