<s>[INST] 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
 [/INST] 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
$$f(\frac{4}{5}\sqrt{5})=25$$
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