Find minimum of function and minimum point 
We have a function $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$ the task to find minimum of this function and point where is minimum. 

I tried to do by mark $t=x+\frac{1}{x}$ and then simplify this function and after that find derivative etc., but it is such a long solution. The professor, in the class, said that there is very short way to do this exercise, maybe somebody know it?

Why is that substitution bad? We get $${t^6-(t^6-6t^4+9t^2-2)\over t^3+(t^3-3t)} = {6t^4-9t^2+2\over 2t^3-3t}$$
$$ = 3t+{2\over 2t^3-3t}$$
Derivative of that is $$3-{12t^2-6\over (2t^3-3t)^2}$$
So we want to solve $$4t^6-12t^4+9t^2= 4t^2-2$$i.e.
$$4t^6-12t^4+5t^2+ 2=0$$
Write $s=t^2$, so we have $$4s^3-12s^2+5s+2=0$$
We see that we have zeroes in following intervals $(-1,0)$, $(0,1)$ and $(2,3)$. Since $$s = (x+{1\over x})^2= x^2+2+{1\over x^2} \geq 2+2 =4$$ we see that this equation has no solution.
