How find this $\frac{3x^3+125y^3}{x-y}$ minimum value 
let $x>y>0$,and such $xy=1$, find follow minimum of the value
  $$\dfrac{3x^3+125y^3}{x-y}$$

My idea: let $x=y+t,t>0$
then
$$\dfrac{3x^3+125y^3}{x-y}=\dfrac{3(y+t)^3+125y^3}{t}=3t^2+3yt+3y^2+\dfrac{128y^3}{t}$$
and $$(y+t)y=1$$
I think this can use AM-GM inequality.But I can't.
Thank you very much

First note that $x>1$ and simplify like this:
$$\frac{3x^3+125y^3}{x-y}=\frac{3x^6+125x^3y^3}{x^4-x^3y}=\frac{3x^6+125}{x^4-x^2}$$
Set $t=x^2$. Now we want to find $a\ge0$ such that
$$\frac{3t^3+125}{t^2-t}\ge a\Longleftrightarrow3t^3+at+125\ge at^2$$
for $t>1$ where the equality is possible.
Note that we could use AM-GM inequality on $t^3+t^3+t^3+at+5^3$, but that equality is possible only if all terms are equal, i.e. $t^3=at=5^3\Longrightarrow t=5, a=5^2$.
Luckily this gives us exactly what we wanted:
$$t^3+t^3+t^3+5^2t+5^3\ge5\sqrt[5]{t^{10}5^5}=5^2t^2$$
Therefore $a=25$ is the minimum and it's possible only for $t=5\Longleftrightarrow x=\sqrt5, y=\dfrac1{\sqrt5}$.
