# Minimum Value of $\frac{x^2}{x-9}$ using AM-GM inequality

I've been trying to find the minimum value of $\frac{x^2}{x-9}$ when x>9 using AM-GM inequality but am unable to do so. The problem is trivial using calculus but I would like to see it done using AM-GM. I am aware that the answer is $36$.

• Check your equation. The left-hand limit of the linked function is $-\infty$ while the right-hand limit is $\infty$. There is no minimum value. Perhaps you mean on a specific interval? Jan 4 '16 at 3:29
• Ah, sorry about that. I've edited the question to only consider when x>9 Jan 4 '16 at 3:36

Let $t=x-9$, then $$\frac{x^2}{x-9}=\frac{(t+9)^2}{t} = t+\frac{81}{t}+18 \ge 2 \sqrt{81}+18=36$$
• @JasonG. Take note that the equality holds if and only if $t=\frac{81}{t}=18$, which is impossible. Jan 5 '16 at 1:20
• It means that $\ge 34$ is always true, but you cannot get to this point, so it makes no sense to this question. Jan 5 '16 at 9:17