If $x > y > 0$, find the minimum of $x + \frac{8}{(y)(x-y)}$.
I have just introduced myself to inequalities (mainly AM-GM) and am trying to use the fact that the minimum of a sum is when both terms are equal, but only when their product is fixed. However I am unable to prove that the product of these two are constant.
Let $x=y+a$.
Hence, by AM-GM $$x+\frac{8}{y(x-y)}=a+y+\frac{8}{ay}\geq a+\frac{4\sqrt2}{\sqrt{a}}=$$ $$=a+\frac{2\sqrt2}{\sqrt{a}}+\frac{2\sqrt2}{\sqrt{a}}\geq3\sqrt[3]{a\left(\frac{2\sqrt2}{\sqrt{a}}\right)^2}=6.$$ The equality occurs for $a=2$, $x=4$ and $y=2$, which says that $6$ is a minimal value.