If $x>4$, what is the minimum value of $\frac{x^4}{(x-4)^2}$. If $x>4$, what is the minimum value of $\frac {x^4}{(x-4)^2}$ ?
I have tried using AM-GM Inequality here by letting $y=x-4$ and ended up getting $224$ but that does not seem to be the correct answer. I find out by trial and error that the minimum value is when $x=8$ which is $256$. Are there any better ways to solve for the minimum value other than trial and error?

Let $p=x-4>0$. By AM-GM inequality,
$$
\frac{x^4}{(x-4)^2}
=\frac{(p+4)^4}{p^2}
\ge\frac{\left(2\sqrt{4p\,}\right)^4}{p^2}
=256
$$
and equality holds when $p=4$ or $x=8$.
