3. Find the smallest possible value of the expression

$$
\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right)
$$

where $x, y, z$ are non-zero real numbers.
Solution. Note that the signs of all six numbers $\frac{x y}{z}, \frac{z x}{y}$, etc., are the same. If all of them are negative, then replace the numbers $x, y, z$ with their absolute values. As a result, each term ($\frac{x y}{z}$, etc.) will change its sign. The modulus of each bracket will remain the same, but the sign will change, so the product of the two brackets will remain the same. Therefore, any value taken by the expression is also taken when $x, y, z$ are positive.

For positive values of $x, y, z$, we use the inequality of the arithmetic mean and the geometric mean. We get:

$\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right) \geqslant 3 \sqrt{\frac{x y}{z} \cdot \frac{z x}{y} \cdot \frac{y z}{x}} \cdot 3 \sqrt{\frac{x}{y z} \cdot \frac{y}{z x} \cdot \frac{z}{x y}}=9 \sqrt{\frac{(x y z)^{2}}{x y z} \cdot \frac{x y z}{(x y z)^{2}}}=9$.

Obviously, the value 9 is achieved, for example, when $x=y=z=1$.