12. (22nd All-Soviet Union Mathematical Olympiad) Given $x, y, z \in \mathbf{R}^{+}$, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $s=\frac{y z}{x}+$ $\frac{z x}{y}+\frac{x y}{z}$.
12. Given $x^{2}+y^{2}+z^{2}=1$, and $x, y, z \in \mathbf{R}^{+}$, we can set $x=\cos \alpha \cdot \cos \beta, y=\cos \alpha \cdot \sin \beta, z=\sin \alpha$, where $\alpha, \beta \in(0$, $\left.\frac{\pi}{2}\right)$, then we have $s=\frac{\sin \alpha \cdot \sin \beta}{\cos \beta}+\frac{\sin \alpha \cdot \cos \beta}{\sin \beta}+\frac{\cos ^{2} \alpha \cdot \sin \beta \cdot \cos \beta}{\sin \alpha}=\frac{2}{\sin 2 \beta} \cdot \sin \alpha+\frac{1-\sin ^{2} \alpha}{2 \sin \alpha} \cdot \sin 2 \beta=\left(\frac{2}{\sin 2 \beta}-\right.$ $\left.\frac{\sin 2 \beta}{2}\right) \sin \alpha+\frac{\sin 2 \beta}{2 \sin \alpha}=\frac{4-\sin ^{2} 2 \beta}{2 \sin 2 \beta} \cdot \sin \alpha+\frac{\sin 2 \beta}{2 \sin \alpha} \geqslant 2 \cdot \sqrt{\frac{4-\sin ^{2} 2 \beta}{2 \sin 2 \beta} \cdot \sin \alpha \cdot \frac{\sin 2 \beta}{2 \sin \alpha}}=\sqrt{4-\sin ^{2} 2 \beta} \geqslant \sqrt{3}$. When and only when $\sin 2 \beta=1$ and $\sin \alpha=\frac{\sqrt{3}}{3}, \cos \alpha=\frac{\sqrt{6}}{3}$, $s$ achieves its minimum value $\sqrt{3}$, at which point it is easy to see that $x=y=z=\frac{\sqrt{3}}{3}$.