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.

Rephrased:
$y:=x-9,$ $y\in (0,\infty)$;
$f(y)=\dfrac{(y+9)^2}{y} =$
$y+18+\dfrac{81}{y}=$
$(y^{1/2}-\dfrac{9}{y^{1/2}})^2+18+18 \ge 36;$
Equality for
$y^{1/2}=\dfrac{9}{y^{1/2}}$,
$y=9.$
