Let $x, y, z$ be real numbers and $x \geq y \geq z \geq \frac{\pi}{12}$. If $x+y+z$ $=\frac{\pi}{2}$, then…
Let $x, y, z$ be real numbers and $x \geq y \geq z \geq \frac{\pi}{12}$. If $x+y+z$ $=\frac{\pi}{2}$, then the minimum value of $\cos x \cdot \sin y \cdot \cos z$ is
$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{6}$
$\frac{1}{8}$
Solution
Given $x+y+z=\frac{\pi}{12}$ and $x \geq y \geq z \geq \quad \frac{\pi}{12}$
Now, Take $\cos \mathrm{x}$ siny $\cos \mathrm{z}$.
$
\begin{aligned}
& \frac{1}{2}(2 \cos x \sin y \cos z)=\frac{1}{2}(\cos x(\sin (y+z)+\sin (y-z))) \\
& \geq \frac{1}{2}(\cos x+\sin y+2)
\end{aligned}
$
Here, $\mathrm{y}+\mathrm{z}=\frac{\pi}{2}-\mathrm{x}$,
$
\geq \frac{1}{2}\left(\cos x \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 $\frac{1}{8}$