Find minimum of the function using AM-GM 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?

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 
$x+ \frac{8}{x} >= 2\sqrt{x*\frac{8}{x}} = 4\sqrt{2}$
