TITLE: Minimum Value of $\frac{x^2}{x-9}$ using AM-GM inequality
QUESTION [4 upvotes]: 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 $36$.

REPLY [1 votes]: wangjiezhe gave a pretty nice and simple solution!
I want to share one more alternate method other than A.M.$\geq G,M.$
For any $x>9,$ we have
$$
\frac{x^{2}}{x-9}-36=\frac{x^{2}-36 x+324}{x-9}=\frac{(x-18)^{2}}{x-9} \geqslant 0 \Rightarrow \frac{x^{2}}{x-9}\geq36
$$
Therefore it attains its minimum 36 when $x=18.$
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