Find minimum of the function using AM-GM

Question

Problem: Find the minimum of the function $f(x,y)=x + \frac{8}{y(x-y)}$, where $x>y>0$ using AM-GM.
My attempt:
$$f(x,y)=2\cdot \frac{x+\frac{8}{y(x-y)}}{2} \ge 2 \sqrt{\frac{8x}{y(x-y)}}$$ But here nothing cancels and I couldn't get a constant. I also tried partial fraction, but didn't work either. $$f(x,y)=x+\frac{8}{xy}-\frac{8}{\color{red}{{x(y-x)}}}$$ I think the issue is with the denominator in red. Any hint please?

Answer 1

Write $z=x-y>0$ then we have, by AM-GM $$f =z+y+{8\over yz} \geq 3\sqrt[3]{yz{8\over yz}} =6$$ with equality iff $z=y=2$. So $x=4$.

Answer 2

First off $x= \frac{y+ (x-y)}{2} >= \sqrt{y(x-y)}$
AM-GM was used for the first time. So now we have $x+ \frac{8}{\sqrt{y(x-y)}} >= x+ \frac{8}{x}$
Then one uses AM-GM inequality for the second time to conclude that
$x+ \frac{8}{x} >= 2\sqrt{x*\frac{8}{x}} = 4\sqrt{2}$