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

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 $x=\dfrac 23$).

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    Why do you think that $9y+\frac{4}{y}$ is minimized when it is $0$? – 2017-01-11
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    @SachpazisStelios Since you usually minimize things by setting them equal to zero... I just learned the AM-GM Inequality like, 15 minutes ago. Can you not to this? – 2017-01-11
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    You set the derivative to zero, not the thing itself. – 2017-01-11
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    @Frank: You have AM=GM when all terms are equal and not when $9y+\frac{4}{y}=0$. Take a look in my answer. – 2017-01-11

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Actually, $(4)$ is minimized with the LHS is equal to $12$, since it is $\geq 12$. We get $$9y + \dfrac{4}{y} = 12$$ is true if and only if $$9y^2 + 4 = 12y$$ or $$9y^2 - 12y + 4 = 0\text{.}$$ Apply the quadratic formula to get $$y = \dfrac{12 \pm \sqrt{12^2-4(9)(4)}}{2(9)} = \dfrac{12 \pm \sqrt{144-144}}{18} = \dfrac{12}{18} = \dfrac{2}{3}\text{.}$$

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Hint: you can show it as $$\frac { 9x^{ 2 }\sin ^{ 2 } x+4 }{ { x\sin x } } =\frac { \left( 3x\sin { x } -2 \right) ^{ 2 }+12x\sin { x } }{ x\sin x } =\frac { \left( 3x\sin { x } -2 \right) ^{ 2 } }{ x\sin x } +12$$

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    Interesting way! I did not even see that! – 2017-01-11
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What you did wrong is that you get AM=GM when all term are equal and not when $9y+\frac{4}{y}=0$.

Then you get $12$ as a minimum when

$$9y=\frac{4}{y}\rightarrow y=\pm \frac{2}{3}$$

but $y \ge 0$ then $y=2/3$.

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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 $y = 2/3$.