12. Let $a, b, c$ be positive constants, and $x^{2}+y^{2}+z^{2}=1$, find
$$f(x, y, z)=\sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}+\sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}}+\sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}}$$

the maximum and minimum values. (1999 China National Training Team Problem)
12. By Cauchy-Schwarz inequality,
$$\left(a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqslant\left(a x^{2}+b y^{2}+c z^{2}\right)^{2}$$

Therefore, $\square$
$$\sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}} \geqslant a x^{2}+b y^{2}+c z^{2}$$

Similarly,
$$\begin{array}{l}
\sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}} \geqslant a y^{2}+b z^{2}+c x^{2} \\
\sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}} \geqslant a z^{2}+b x^{2}+c y^{2}
\end{array}$$

Adding the three inequalities, we get
$$\begin{array}{l}
\sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}+\sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}}+\sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}} \geqslant \\
a+b+c
\end{array}$$

By the AM-GM inequality,
$$\begin{array}{c}
2 \sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}} \sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}} \leqslant \\
a^{2}\left(x^{2}+y^{2}\right)+b^{2}\left(y^{2}+z^{2}\right)+c^{2}\left(z^{2}+x^{2}\right) \\
2 \sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}} \sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}} \leqslant \\
a^{2}\left(y^{2}+z^{2}\right)+b^{2}\left(z^{2}+x^{2}\right)+c^{2}\left(x^{2}+y^{2}\right) \\
2 \sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}} \sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}} \leqslant \\
a^{2}\left(z^{2}+x^{2}\right)+b^{2}\left(x^{2}+y^{2}\right)+c^{2}\left(y^{2}+z^{2}\right) \\
\left(a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}\right)+\left(a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}\right)+\left(a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}\right)= \\
\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)
\end{array}$$

Adding the four inequalities and noting that $x^{2}+y^{2}+z^{2}=1$, we get
$$f^{2}(x, y, z) \leqslant 3\left(a^{2}+b^{2}+c^{2}\right)$$

Therefore,
$$\begin{aligned}
\sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}+ & \sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}}+\sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}} \leqslant \\
& \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}
\end{aligned}$$

Thus, the maximum value of $f(x, y, z)$ is $\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}$, and the minimum value is $a+b+c$.