Let x, y, z be real numbers and x ≥ y ≥ z ≥ (π)/(12) . If...

Given $x+y+z=\frac{\pi}{12} and$x \geq y \geq z \geq \quad \frac{\pi}{12}$Now, Take$\cos \mathrm{x}$\sin y$\cos \mathrm{z}$.$$\begin{aligned}& \frac{1}{2}(2 \cos x \sin y \cos z)=\frac{1}{2}(\cosx(\sin (y+z)+\sin (y-z))) \\ & \geq \frac{1}{2}(\cosx+\sin y+2) \end{aligned}$$Here,$\mathrm{y}+\mathrm{z}=\frac{\pi}{2}-\mathrm{x}$, \geq$\frac{1}{2}\left(\cosx \sin \left(\frac{\pi}{2}-x\right)\right) \geq $\frac{1}{2} \cos ^2 x . If we take minimum value of$y=z=\frac{\pi}{12}$, then$x=\frac{\pi}{3}$So,$\frac{1}{2} \cos ^2 \mathrm{x}=\frac{1}{2}\left(\cos \frac{\pi}{3}\right)^2=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}$Therefore, Minimum value of the given expression is$\frac{1}{8}$