If $x>4$, what is the minimum value of $\frac{x^4}{(x-4)^2}$.

Question

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?

Answer 1

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 $x=8$.

Answer 2

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 $x=8$.

Answer 3

If cooking up AM-GM doesn't strike you immediately, here is an alternative :

Let $y = \dfrac{x^2}{x-4} \Rightarrow x^2-xy+4y = 0$

$x$ as a function of $y$ is defined when discriminant $y^2 - 16y \ge 0$.