Minimum value of $\frac{\sin x}{4} + \frac{1}{\sin x}$? $x \in (0, \pi/2)$

Question

Question. What is the minimum value of $$ \frac{\sin x}{4} + \frac{1}{\sin x}$$ subject to $x \in (0, \pi/2)$?

Applying AM-GM inequality, we get

$$ \frac{\sin x}{4} + \frac{1}{\sin x} \geq 1. $$

However, for the expression to be equal to $1$, $\sin x$ must be $2$, which is not possible. Moreover, on plotting the graph, the minimum value is $\frac{5}{4}$.

Please explain.

Answer 1

Another way to solve the exercise without using derivatives.

For any $\,x\in\left(0,\dfrac{\pi}2\right]\,$ it results that

$\begin{align}\dfrac{\sin x}4+\dfrac1{\sin x}&=\sin x+\dfrac1{\sin x}-\dfrac34\sin x=\\[3pt]&=\left(\!\sqrt{\sin x}-\dfrac1{\sqrt{\sin x}}\!\right)^{\!2}\!+2-\dfrac34\sin x\geqslant\\[3pt]&\geqslant2-\dfrac34\sin x\geqslant2-\dfrac34=\dfrac54\;.\end{align}$

Moreover ,

$\dfrac{\sin x}4+\dfrac1{\sin x}=\dfrac54\iff\sin x=1\iff x=\dfrac{\pi}2\;.$

Consequently, the minimum value of $\,\dfrac{\sin x}4+\dfrac1{\sin x}\,$ subject to $\,x\in\left(0,\dfrac{\pi}2\right]\,$ is $\,\dfrac54\,,\,$ but the function $\,\dfrac{\sin x}4+\dfrac1{\sin x}\,$ does not have a minimum on the interval $\left(0,\dfrac{\pi}2\right),\,$ in fact it only has the infimum (the greatest lower bound) which is $\,\dfrac54\,.$

Answer 2

Write $f(x) = \dfrac x4 + \dfrac 1x$. Then $f'(x) = \dfrac 14 - \dfrac 1{x^2}$ which is negative when $0 < x < 2$.

Thus $f$ is decreasing on the interval $(0,1]$, so its minimum value on that interval occurs at $x=1$ and the minimum value of $f$ is $f(1) = \dfrac 54$.

Finally observe $x \in (0,\pi/2)$ implies $\sin x \in (0,1)$.