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


My idea: let $x=y+t,t&gt;0$
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 $x&gt;1$ 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 $t&gt;1$ 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 $t=5\Longleftrightarrow x=\sqrt5, y=\dfrac1{\sqrt5}$.
continuation:
let $x-y=t$, $xy=1$, we get:
$$x=\frac{1}{2}t+\frac{1}{2}\sqrt{t^2+4},\\y=-\frac{1}{2}t+\frac{1}{2}\sqrt{t^2+4}$$
then we have
$$f=\frac{3x^3+125y^3}{x-y}\\=\frac{-61t^3+64t^2\sqrt{t^2+4}-183t+64\sqrt{t^2+4}}{t}$$
we calculate the differential:
$$g=\frac{df}{dt}=-\frac{2(61t^3\sqrt{t^2+4}-64t^4-128t^2+128)}{t^2\sqrt{t^2+4}}$$
let $g=0$, we get the good solution:
$$t=\frac{4}{5}\sqrt{5}$$
let $t=\frac{4}{5}\sqrt{5}$, we have
$$f(\frac{4}{5}\sqrt{5})=25$$