3.1. For what greatest $a$ is the inequality $\frac{\sqrt[3]{\operatorname{tg} x}-\sqrt[3]{\operatorname{ctg} x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}>\frac{a}{2}$ satisfied for all permissible $x \in\left(\frac{3 \pi}{2} ; 2 \pi\right)$? Round the answer to the nearest hundredth if necessary.
Answer: 4.49 (exact value: $4 \sqrt[6]{2}$).

Solution. Transform the expression on the left side of the inequality as follows:

$$
\frac{\sqrt[3]{\operatorname{tg} x}-\sqrt[3]{\operatorname{ctg} x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}=\frac{\sqrt[3]{\sin ^{2} x}-\sqrt[3]{\cos ^{2} x}}{\sqrt[3]{\sin x \cos x}(\sqrt[3]{\sin x}+\sqrt[3]{\cos x})}=\frac{\sqrt[3]{\sin x}-\sqrt[3]{\cos x}}{\sqrt[3]{\sin x \cos x}}=\frac{1}{\sqrt[3]{\cos x}}+\frac{1}{\sqrt[3]{-\sin x}}
$$

if $\cos x \neq -\sin x$. For $x \in \left(\frac{3 \pi}{2} ; 2 \pi\right)$, both terms are positive, so by the inequality between the arithmetic mean and the geometric mean, we get

$$
\frac{1}{\sqrt[3]{\cos x}}+\frac{1}{\sqrt[3]{-\sin x}} \geqslant \frac{2}{\sqrt[6]{-\sin x \cos x}}=\frac{2 \sqrt[6]{2}}{\sqrt[6]{-\sin 2 x}} \geqslant 2 \sqrt[6]{2}
$$

with equality holding only at $x=\frac{7 \pi}{4}$, but this point is not valid for the given expression. The original expression for $x$ close to $\frac{7 \pi}{4}$ takes values close to $2 \sqrt[6]{2}$. Therefore, the desired value of $a$ is $4 \sqrt[6]{2} \approx 4.49$.