2. Given numbers $x, y \in\left(0, \frac{\pi}{2}\right)$. Find the maximum value of the expression

$$
A=\frac{\sqrt[4]{\sin x \sin y}}{\sqrt[4]{\operatorname{tg} x}+\sqrt[4]{\operatorname{tg} y}}
$$
Answer: $\frac{\sqrt[4]{8}}{4}$.

Solution. Note that $\sqrt[4]{a}+\sqrt[4]{b} \leqslant \sqrt[4]{8(a+b)}$ for any $a, b \geqslant 0$, since

$$
(\sqrt[4]{a}+\sqrt[4]{b})^{4} \leqslant(2(\sqrt{a}+\sqrt{b}))^{2} \leqslant 8(a+b)^{4}
$$

Applying the inequality for the harmonic mean and the arithmetic mean, we get

$$
\begin{aligned}
A \leqslant \frac{1}{4} \sqrt[4]{\sin x \sin y}(\sqrt[4]{\operatorname{ctg} x}+\sqrt[4]{\operatorname{ctg} y}) & =\frac{1}{4}(\sqrt[4]{\sin x \cos y}+\sqrt[4]{\cos x \sin y}) \leqslant \\
& \leqslant \frac{\sqrt[4]{8}}{4} \sqrt[4]{\sin x \cos y+\cos x \sin y}=\frac{\sqrt[4]{8}}{4} \sqrt[4]{\sin (x+y)} \leqslant \frac{\sqrt[4]{8}}{4}
\end{aligned}
$$

Equality is achieved when $x=y=\frac{\pi}{4}$.