Minimum value of $\frac{x^2}{x-9}$
I was asked to use the AM-GM inequality to solve this, I was thinking that I should express it into 2 fractions, and so I did, I tried to express $x^2$ as $\frac{(x-9)^2 +18x -81 }{x-9}$, though I couldn't find an answer.
Hint:
Set $x-9=h$ $$\dfrac{x^2}{x-9}=\dfrac{(h+9)^2}h=h+\dfrac{9^2}h+18$$
If $h>0\iff x-9>0$ $$\dfrac{h+\dfrac{9^2}h}2\ge\sqrt{h\cdot\dfrac{9^2}h}=?$$