Problem 5. Solve the equation: $4\left(x^{4}+3 x^{2}+3\right)\left(y^{4}-7 y^{2}+14\right)=21$.
Answer: $\left(0 ; \pm \sqrt{\frac{7}{2}}\right)$.

Solution. Note that $x^{4}+3 x^{2}+3 \geqslant 3$ for each $x$, and $y^{4}-7 y^{2}+14=\left(y^{2}-7 / 2\right)^{2}+7 / 4 \geqslant 7 / 4$ for each $y$. Therefore, the left side of the equation is greater than or equal to $4 \cdot 3 \cdot 7 / 4=21$, and equality is achieved only when $x=0$ and $y^{2}=7 / 2$. This gives the answer.