Minimum of a function by inequality

Question

No answers please, hints only!

I am asked to use the fact that $$\cfrac{a_1+\cdots+a_n}{n}\ge \sqrt[n]{a_1\cdots a_n}, n\in \mathbb N \\ \text{with equality if and only if } a_1=a_2=\cdots=a_n$$ To find the minimum value (and when this value occurs) of the function $$f(x)=\cfrac{9x^2(\sin^2x)+4}{x\sin x}, \hspace{0.5cm} x\in (0, \pi)$$

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I've decided to set $f(x)\ge K$ to make it easier to use this inequality. \begin{align*} \cfrac{9x^2(\sin^2x)+4}{x\sin x} &\ge K \\ \cfrac{9x^2(\sin^2x)+4}{x} &\ge K\sin x \\ \cfrac{\biggr(3x\sin x\biggr)^2+2^2}{x} &\ge K\sin x\end{align*} And then this is the point I didn't know how to continue, and I'm not sure if I even have the right approach to this.

Answer 1

Hint: Multiplying top and bottom by $\frac2{x \sin(x)}$ gives $$f(x) = \frac{18x \sin(x) + \frac8{x \sin(x)}}{2} $$

Answer 2

HINT:

Consider the function $$g(x)=\frac{x\sin(x)f(x)}2=\frac{(3x\sin x)^2+2^2}2$$

Note that we can apply AM-GM (your fact) to get a lower bound on $g(x)$. Using your fact, you can also show that when $(3x\sin x)^2=2^2$, then $g(x)$ equals the lower bound.

Now multiply your inequality by $\frac2{x\sin x}$ to get a lower bound on $f(x)$.