10. $\alpha, \beta, \gamma$ are all acute angles, and $\cos ^{2} \alpha+\cos ^{2} \beta+\cos \gamma^{2}=1$, find the minimum value of $\tan \alpha \cdot \tan \beta \cdot \tan \gamma$.
10. Since $(\tan \alpha \cdot \tan \beta \cdot \tan \gamma)^{2}=\tan ^{2} \alpha \tan ^{2} \beta \tan ^{2} \gamma=\left(\frac{1}{\cos ^{2} \alpha}-1\right)\left(\frac{1}{\cos ^{2} \beta}-1\right)\left(\frac{1}{\cos ^{2} \gamma}-\right.$ $1)=\frac{1-\cos ^{2} \alpha}{\cos ^{2} \alpha} \cdot \frac{1-\cos ^{2} \beta}{\cos ^{2} \beta} \cdot \frac{1-\cos ^{2} \gamma}{\cos ^{2} \gamma}$. Given $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$, hence $1-\cos ^{2} \alpha=$ $\cos ^{2} \beta+\cos ^{2} \gamma \geqslant 2 \cos \beta \cos \gamma$. Similarly, $1-\cos ^{2} \beta=\cos ^{2} \alpha+\cos ^{2} \gamma \geqslant 2 \cos \alpha \cos \gamma, 1-\cos ^{2} \gamma=$ $\cos ^{2} \alpha+\cos ^{2} \beta \geqslant 2 \cos \alpha \cos \beta$. Therefore, $(\tan \alpha \cdot \tan \beta \cdot \tan \gamma)^{2} \geqslant 8, \tan \alpha>0, \tan \beta>0, \tan \gamma>0$, hence $\tan \alpha \cdot \tan \beta \cdot \tan \gamma \geqslant 2 \sqrt{2}$. When and only when $\cos \alpha=\cos \beta=\cos \gamma=\frac{\sqrt{3}}{3}$, $\tan \alpha \cdot \tan \beta \cdot \tan \gamma$ achieves its minimum value of $2 \sqrt{2}$.