# 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?

• I'm guessing derivatives are off the table here. – Arthur Aug 24 '19 at 8:17
• If it's easier, you may try finding the minimum of sqrt of the given expression – AgentS Aug 24 '19 at 8:19

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 $$x=8$$.
Let $$y = \dfrac{x^2}{x-4} \Rightarrow x^2-xy+4y = 0$$
$$x$$ as a function of $$y$$ is defined when discriminant $$y^2 - 16y \ge 0$$.
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 $$x=8$$.