What is the minimum value of $\dfrac {9x^2\sin^2 x+4}{x\sin x}$ 
Question: What is the minimum value of $$\dfrac {9x^2\sin^2x+4}{x\sin x}\tag1$$For $0<x<\pi$.


I solved this, but somewhere, I did something wrong. My work is as follows:
First, set $y=x\sin x$. Therefore,
$$\begin{align*}\dfrac {9y^2+4}{y}=\dfrac 4y+9y\tag2\end{align*}$$
And by the AM-GM Inequality, we have$$\begin{align*}\dfrac {9y+\dfrac 4y}2 & \geq\sqrt{\dfrac 4y\cdot 9y}\tag3\\9y+\dfrac 4y & \geq12\tag4\end{align*}$$
$(4)$ is minimized only when the RHS is equal to zero. So$$9y+\dfrac 4y=0\implies 9y^2=-4\implies y^2=-\dfrac 49\tag5$$
However, if I square root both sides, I get an imaginary value. Where did I go wrong? The book says the solution is $12$ (received by plugging in $x=\dfrac 23$).

You already noted that
$$9y + \frac{4}{y} \geq 12$$ 
so if the LHS were exactly $12$, it would certainly be minimal!
If equality is to be attained, you have
$$9y + \frac{4}{y} - 12 = 0,$$
multiplying by $y$ gives
$$9y^2 - 12y + 4 = (3y - 2)^2 = 0$$
Hence $y = 2/3$.
