Showing that $\frac{9x^2\sin^2x + 4}{x\sin x}$ has a local minimum for $0 < x < \pi$

Question

Assume that only pen/pencil and paper may be used for this question. (i.e., no numeric solvers.)

Demonstrate that $$\dfrac{9x^2\sin^2x + 4}{x\sin x}$$ has a local minimum for $0 < x < \pi$.

My attempts. Let $y = x\sin x$ (which we observe is $> 0$ over the given interval) and $$f(y) = \dfrac{9y^2 + 4}{y} = 9y + \dfrac{4}{y}\text{.}$$

We observe that, letting $y = x\sin x$, that if $f(y) = 9y + \dfrac{4}{y}$, then $$f^{\prime}(y) = y^{\prime}\left(9 - \dfrac{4}{y^2} \right)\text{.}$$ Clearly this equation's roots are when either $y^{\prime} = 0$ or $9- \dfrac{4}{y^2} = 0$. I focus on $y^{\prime}$. We have $$y^{\prime} = \sin x + x\cos x\text{.}$$ Since $\sin x > 0$ and $x > 0$, the sign is completely determined by $\cos x$. Obviously $y^{\prime} > 0$ for $0 < x < \dfrac{\pi}{2}$; we then observe that $$y^{\prime}\left(\dfrac{3\pi}{4} \right) < 0 $$ so by the intermediate value theorem, $y^{\prime}$ is positive from $(0, c)$ and negative from $(c, \pi)$ for some $0 < c < \pi$.

I am not sure how to deal with analyzing $9 - \dfrac{4}{y^2}$ by hand, and I'm not satisfied with my analysis of $y^{\prime}$.

Answer 1

You can argue using AM-GM:

You have for $y= x\sin x$

$$9y+\frac 4y \stackrel{AM-GM}{\geq} 2\sqrt{9y\cdot \frac 4y}=12$$

Equality is achieved for $$9y=\frac 4y \stackrel{y>0}{\Leftrightarrow } y=\frac 23,$$ which lies in the image of $(0,\pi)$ under $x\sin x$.

Answer 2

$9 - \frac {4}{y^2} = 0\\ 9y^2 - 4 = 0\\ y = \pm \frac {2}{3}$

Since $y(0) = 0, y(\pi) = 0,$ and $y(\frac \pi 2) = \frac {\pi}{2} > \frac 23$ there exists at least one zero for $9-\frac {4}{y^2}$ in $(0,\frac {\pi}{2})$ and one in $(\frac {\pi}{2}, \pi)$

We know that $y' > 0$ for $x\in (0,\frac {\pi}{2}],$ so lets look for a minimum in this interval.

$y$ is strictly increasing in this interval.

$\frac {d}{dx} \frac {9x^2 + 4}{x\sin x} < 0$ when $y < \frac {2}{3}$ and greater than $0$ when $y > \frac {2}{3}$ thus there is a minimum when $y = \frac {2}{3}$

Answer 3

You technically don’t need to know that much about the functions.

You know $f(x)=x\sin x$ on $(0,\pi)$ has range $(0,M]$ for some $M>0.$ This is because the values are positive and $x\sin x\to 0$ as $x\to0^+$ and $x\to\pi^-.$

Using standard calculus, you can show that $g(u)=9u+\frac{1}{4u}$ has a continuous increasing derivative on $\mathbb R^+$ with some $g’(u)<0.$ This means that either $g$ is always decreasing, or is decreasing until some point, and then increasing after that. Let $u_0$ be the value that minimizes $g(u_0),$ or $u_0=+\infty$if it is strictly decreasing.

Now, let $h(x)=g(f(x)).$ Then if $M\geq u_0,$ there is some $x_0$ such that $f(x_0)=u_0,$ and thus $x_0$ is an absolute minimum for $h(x),$ not just a local minimum.

If $M<u_0,$ then the absolute minimum of $g$ on $[0,M]$ is when $u=M.$ Pick $x_0$ such that $f(x_0)=M.$ Then $h(x_0)$ is the absolute minimum of $h$ on $(a,b).$

So we get a general rule:

If $f(x)$ is continuous on $[a,b]$ such that $f(a)=f(b)=0$ and $f(x)>0$ for $x\in(a,b)$ and $g(u)$ is a continuous function on $\mathbb R^+$ which is either:

1. Strictly decreasing, or
2. Decreasing for $u<u_0$ and increasing for $u>u_0,$ for some $u_0>0.$

Then $g(f(x))$ has a local minimum on $(a,b).$

The key is, you don’t need the functions to be differentiable, only continuous. We might use differentiation of the particular $g(u)$ to show it satisfies 2., but we might do so in another way, such as if $g(u)=|u-u_0|.$