Functional analysis problem.

Question

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.

Answer 1

You're almost there: $\;x^2-ax+9a\ge0\iff \Delta\le0\;$ ,where $\;\Delta:=\,$ the discriminant of the given quadratic. Can you continue from here? Without derivatives and stuff: just basic algebra and basic geometry.

Answer 2

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

$$x-9 = \frac{81}{x-9} \stackrel{x>9}{\Leftrightarrow} x=18$$

Answer 3

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

$y^{1/2}=\dfrac{9}{y^{1/2}}$,

$y=9.$