Functional analysis problem. Find the smallest value of $f(x) = \frac{x^2}{x-9}$ on the interval $(9, +\infty)$.
We should basically find the biggest $a$ such that $\frac{x^2}{x-9} \geq a$. We can multiply both sides by $x-9$ since it's positive and than we get $x^2-ax+9a \geq 0.$ I don't know how to proceed.

Here is a purely algebraic way using the inequality between arithmetic and geometric mean (AM-GM):
For $x>9$ we have
\begin{eqnarray*} \frac{x^2}{x-9}
& = & \frac{x^2-81+81}{x-9} \\
& = & x+9 + \frac{81}{x-9} \\
& = & x-9 + \frac{81}{x-9} + 18 \\
& \stackrel{AM-GM}{\geq} & 2\sqrt{81} + 18\\
& = & 36
\end{eqnarray*}
Equality holds if and only if
$$x-9 = \frac{81}{x-9} \stackrel{x>9}{\Leftrightarrow} x=18$$
