3. Find the minimum value of the expression $\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}}$ under the conditions $|x|<1,|y|<2$ and $x y=1$.

Answer: 4.
Solution. Using the inequality between the arithmetic mean and the geometric mean, under the given conditions on $x$ and $y$, we obtain

$$
\frac{1}{1-x^{2}}+\frac{4}{4-y^{2}} \geqslant 2 \sqrt{\frac{1}{1-x^{2}} \cdot \frac{4}{4-y^{2}}}=\frac{4}{\sqrt{5-4 x^{2}-y^{2}}}=\frac{4}{\sqrt{1-(2 x-y)^{2}}} \geqslant 4
$$

and all inequalities become equalities if $2 x=y$ and $x y=1$, i.e., $x=\frac{1}{\sqrt{2}}, y=\sqrt{2}$.