Minimize $\frac{x^2}{x-9} $ given that $x > 9$ Its given that $ x > 9 $ and I have to find minima of
$$ y = \frac{x^2}{(x-9) }   $$
I did this using three methods.
Method 1). Let
$$ f(x) = \frac{x^2}{(x-9) }   $$
Using, calculus, we get
$$ f'(x) = \frac{2x}{(x-9)} - \frac{x^2}{(x-9)^2}   $$
Using this, we can find that the critical point is $ x = 18$ and $ f'(x) < 0 $ for $ 9 < x < 18 $ and $ f'(x) > 0 $ for $ x > 18 $. So, using first derivative test, $ f(x) $ has absolute minimum at $x = 18$.  So, we can find that $f(18) = 36 $ is the absolute minimum.
Method 2). Since we have
$$ y = \frac{x^2}{(x-9) }   $$
$$ \Longrightarrow  x^2 - xy +9y = 0 $$
This is a quadratic equation. Now $x > 9$, so it follows that $ y > 0 $.  Now the discriminant of this quadratic equation is
$$ D = y^2 - 36 y $$
Since roots of this quadratic equation are values of $x$ and since $x > 9$, we know that the roots are real and so  we have discriminant $ D \geqslant 0 $, which means that
$$ y ( y-36) \geqslant 0 $$
Since $ y > 0 $, it follows that $ y \geqslant 36 $. So, 36 is the minimum value of $y$.
Method 3.
Now, we have
$$ \frac{(x-9) }{x^2} = \frac{1}{x} - \frac{9}{x^2}   $$
$$ \frac{(x-9) }{x^2} = \Big( \frac{1}{9} \Big) \Big( \frac{9}{x} \Big) \Big( 1- \frac{9}{x} \Big) $$
We should note that since $ x > 9$, all quantities above are positive. So, we can now use AM-GM inequality
$$ \frac{9}{x} + \Big( 1- \frac{9}{x} \Big)  \geqslant 2 \bigg[ \frac{9}{x}\Big( 1- \frac{9}{x} \Big) \bigg]^{1/2} $$
$$ 1 \geqslant 6 \bigg [ \frac{1}{x} \Big( 1- \frac{9}{x} \Big) \bigg]^{1/2} $$
$$ \therefore \; \bigg[ \frac{1}{x} - \frac{9}{x^2} \bigg ]  \leqslant \frac{1}{36} $$
$$ \therefore \; \frac{(x-9) }{x^2} \leqslant \frac{1}{36} $$
$$ \therefore \; \frac{x^2}{x-9} \geqslant 36 $$
So, the minimum value is $36$
Are these good approaches ?

Yet another method:
$$ \frac{x^2}{x-9}=\frac{(x-18)^2+36x-18^2}{x-9}
=36+\frac{(x-18)^2}{x-9}\ge 36$$
with equality iff $x=18$.
