Determine Minimum Value. 
Find the minimum value of
  $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
  for $x>0$.

When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks
I think differentiation would be really complicated

$$\dfrac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=\dfrac{(x+1/x)^6-(x^3+1/x^3)^2}{(x+1/x)^3+(x^3+1/x^3)}=(x+1/x)^3-(x^3+1/x^3)$$ Therefore your function reduces to $3(x+1/x)$ which has its minimum when $x=1.$
