1983 AIME Problems/Problem 9

Problem

Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$ .

Let $y=x\sin{x}$ .

We can rewrite the expression as $\frac{9y^2+4}{y}=9y+\frac{4}{y}$ .

Since $x>0$ and $\sin{x}>0$ because $0< x<\pi$ , we have $y>0$ .

So we can apply AM-GM:

$9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12$

The equality holds when $9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23$

Therefore, the minimum value is $12$ (when $x\sin{x}=\frac23$ ).