Step 1: Understand the constraints.
We have $ x \geq y \geq z \geq \frac{\pi}{12} $ and $ x + y + z = \frac{\pi}{2} $. We want the minimum value of $ \cos x \sin y \cos z $.
Step 2: Think about when cos x is smallest.
On $ (0, \frac{\pi}{2}) $, the cosine function decreases as the angle increases. So $ \cos x $ is smallest when $ x $ is largest.
Step 3: Find the maximum value of x.
Since $ y \geq z \geq \frac{\pi}{12} $, we have $ y + z \geq \frac{\pi}{6} $. So $ x = \frac{\pi}{2} - (y+z) \leq \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} $. Maximum $ x = \frac{\pi}{3} $.
Step 4: Determine y and z when x = pi/3.
If $ x = \frac{\pi}{3} $, then $ y + z = \frac{\pi}{6} $. Combined with $ y \geq z \geq \frac{\pi}{12} $ and $ y + z = \frac{\pi}{6} $, the only possibility is $ y = z = \frac{\pi}{12} $.
Step 5: Compute the product.
\[ \cos\frac{\pi}{3} \cdot \sin\frac{\pi}{12} \cdot \cos\frac{\pi}{12} = \frac{1}{2} \cdot \sin\frac{\pi}{12} \cdot \cos\frac{\pi}{12} = \frac{1}{2} \cdot \frac{1}{2}\sin\frac{\pi}{6} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \] (using $ 2\sin A\cos A = \sin 2A $ and $ \sin\frac{\pi}{6} = \frac{1}{2} $).
Step 6: State the minimum value.
\[ \boxed{\frac{1}{8}} \]