3. Given numbers $x_{1}, \ldots, x_{n} \in(0,1)$. Find the maximum value of the expression

$$
A=\frac{\sqrt[4]{1-x_{1}}+\ldots+\sqrt[4]{1-x_{n}}}{\frac{1}{\sqrt[4]{x_{1}}}+\ldots+\frac{1}{\sqrt[4]{x_{n}}}}
$$
Answer: $\frac{\sqrt{2}}{2}$.

Solution. By the inequality for the harmonic mean and the arithmetic mean

$$
\frac{1}{\frac{1}{\sqrt[4]{x_{1}}}+\ldots+\frac{1}{\sqrt[4]{x_{n}}}} \leqslant \frac{\sqrt[4]{x_{1}}+\ldots+\sqrt[4]{x_{n}}}{n^{2}}
$$

Note that for any $a_{1}, \ldots, a_{n}$

$$
\left(\sum_{k=1}^{n} a_{k}\right)^{4} \leqslant\left(n \sum_{k=1}^{n} a_{k}^{2}\right)^{2} \leqslant n^{3} \sum_{k=1}^{n} a_{k}^{4}
$$

From here, by the Cauchy inequality

$$
A^{2} \leqslant \frac{\left(\sum_{k=1}^{n} \sqrt[4]{1-x_{k}}\right)^{2} \cdot\left(\sum_{k=1}^{n} \sqrt[4]{x_{k}}\right)^{2}}{n^{4}} \leqslant \frac{\left(\sum_{k=1}^{n} \sqrt[4]{1-x_{k}}\right)^{4}+\left(\sum_{k=1}^{n} \sqrt[4]{x_{k}}\right)^{4}}{2 n^{4}} \leqslant \frac{\sum_{k=1}^{n}\left(1-x_{k}\right)+\sum_{k=1}^{n} x_{k}}{2 n}=\frac{1}{2}
$$

Therefore, $A \leqslant \frac{\sqrt{2}}{2}$. Equality is achieved when $x_{1}=\ldots=x_{n}=\frac{1}{2}$.