1. In the set of real numbers, solve the equation

$$
\left(x^{2}-x+1\right)\left(4 y^{2}+6 y+4\right)\left(4 z^{2}-12 z+25\right)=21
$$
Solution. We have

$$
x^{2}-x+1=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4}
$$

$$
4 y^{2}+6 y+4=4\left(y+\frac{3}{4}\right)^{2}+\frac{7}{4} \geq \frac{7}{4}
$$

and

$$
4 z^{2}-12 z+25=4\left(z-\frac{3}{2}\right)^{2}+16 \geq 16
$$

From (1), (2) and (3) it follows that

$$
\left(x^{2}-x+1\right)\left(4 y^{2}+6 y+4\right)\left(4 z^{2}-12 z+25\right) \geq 16 \cdot \frac{3}{4} \cdot \frac{7}{4}=21
$$

Equality is achieved if and only if equalities are satisfied in (1), (2) and (3). However, the condition for equality is the given equation

$$
\left(x^{2}-x+1\right)\left(4 y^{2}+6 y+4\right)\left(4 z^{2}-12 z+25\right)=21
$$

so the solution to the equation is $x=\frac{1}{2}, y=-\frac{3}{4}, z=\frac{3}{2}$.