# Normal AM-GM inequality

Minimum value of 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 (x-9)^2 +18x -81 /x-9, though i couldn't find an answer.

## 1 Answer

Hint:

Set $x-9=h$ $$\dfrac{x^2}{x-9}=\dfrac{(h+9)^2}h=h+\dfrac{9^2}h+18$$

If $h>0\iff x-9>0$ $$\dfrac{h+\dfrac{9^2}h}2\ge\sqrt{h\cdot\dfrac{9^2}h}=?$$