<s>[INST] 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?
 [/INST] Using AM-GM:
$$\frac {x^4}{(x-4)^2}=\left(\frac{x^2-16+16}{x-4}\right)^2=\left(\color{red}{x-4+\frac{16}{x-4}}+8\right)^2\ge (\color{red}8+8)^2=256,$$
equality occurs for $x=8$.
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