Find the minimum of the function

$$
f(x)=\frac{\cos (x)^{3}}{\sin (x)}+\frac{\sin (x)^{3}}{\cos (x)}
$$

on $] 0, \pi / 2[$.

Theorem 1.70 (Tchebycheff's Inequality). With the same notations as before, if the $a_{i}$ and the $b_{i}$ are arranged in the same order, then

$$
\sum a_{i} b_{i} \geq \frac{1}{n}\left(\sum a_{i}\right)\left(\sum b_{i}\right)
$$

Proof. We use the rearrangement inequality to find the following $n$ relations

$$
\begin{aligned}
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} & \geq a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \\
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} & \geq a_{1} b_{2}+a_{2} b_{3}+\cdots+a_{n} b_{1} \\
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} & \geq a_{1} b_{3}+a_{2} b_{4}+\cdots+a_{n} b_{2} \\
\cdots & \cdots \\
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} & \geq a_{1} b_{n}+a_{2} b_{1}+\cdots+a_{n} b_{n-1}
\end{aligned}
$$

And by adding these relations we obtain the desired inequality.
Let's use this exercise to recall how to determine the minimum of a function $f$ is $m$. This involves two steps: first, show that $f(x) \geq m$ for all $x$, and then find an $x_{0}$ where this minimum is achieved. Consider the function of interest,
$f(x)=\frac{\cos \left(x^{3}\right)}{\sin (x)}+\frac{\sin (x)^{3}}{\cos (x)}$. It is easy to see that the sequences $\left(\cos (x)^{3}, \sin (x)^{3}\right)$ and $(1 / \sin (x), 1 / \cos (x))$ are in the same order. The rearrangement inequality gives

$$
\begin{aligned}
\frac{\cos (x)^{3}}{\sin (x)}+\frac{\sin (x)^{3}}{\cos (x)} \geq f(x) & \geq \frac{\cos (x)^{3}}{\cos (x)}+\frac{\sin (x)^{3}}{\sin (x)} \\
& =\cos (x)^{2}+\sin (x)^{2}=1
\end{aligned}
$$

For the last step, we just note that $f(\pi / 4)=1$, which completes the proof.