Minimum of product of three trigonometric function 
If $x,y,z$ are real numbers with $x\geq y\geq z \geq 15^\circ$ and $x+y+z=90^\circ$, then find the range of $\cos x\cos y\cos z$.

I tried Jensen inequality and AM-GM inequality:
$$3(\cos x\cos y\cos z)^{\frac{1}{3}}\leq \cos x+\cos y+\cos z\leq 3 \cos\bigg(\frac{x+y+z}{3}\bigg)=\frac{3\sqrt{3}}{2},$$
so
$$ \cos x\cos y\cos z\leq \frac{3\sqrt{3}}{8}.$$
Equality holds when $x=y=z=30^\circ$.
Also,$$ \cos x\cos y\cos z=\frac{1}{4}\bigg[\sin(2x)+\sin(2y)+\sin(2x+2y)\bigg].$$
I am unable to find the minimum of $\cos x\cos y\cos z$.

The least value of function 
$$f(x,y) = \cos x\cos y \cos \left(\dfrac\pi2 - x-y\right),$$
or
$$f(x,y) = \cos x\cos y \sin(x+y),$$
can be achieved in the stationary  points of $f(x)$ or in the bounds of the domain.
The stationary points can be defined from the system $f'_x = f'_y = 0,$ or
\begin{cases}
-\sin x \cos y \sin(x+y) + \cos x\cos y\cos(x+y)=0,\\
-\cos x \sin y \sin(x+y) + \cos x\cos y\cos(x+y)=0.
\end{cases}
Subtraction leads to equation
\begin{align}
\sin(x-y)\sin(x+y)=0.
\end{align}
Taking in account the issue conditions, this means
$$y=x,\quad -\sin x \sin 2x + \cos x \cos 2x = 0,\quad x \in[15^\circ, 37.5^\circ],$$
$$\cos 3x =0,\quad y=x,\quad x \in[15^\circ, 37.5^\circ],$$
\begin{align}
x=y=z = 30^\circ,\quad f(30^\circ, 30^\circ) = \dfrac{3\sqrt3}8.
\end{align}
The task is symmetric on $x, y, z,$ so WLOG it is sufficient to consider the bounds $z=15^\circ,\ x+y = 75^\circ,$ and $z= 60^\circ,\ x=y = 15^\circ.$
The first bound gives
$$g(x) = f(x,75^\circ-x) = \cos15^\circ\cos x\cos (75^\circ-x),\quad x\in[15^\circ,\ 60^\circ],$$
$$g(x) = \dfrac12\cos15^\circ(\cos 37.5^\circ + \cos(37.5^\circ - 2x)),$$
with the least value
$$g_m = g(15^\circ) = g(60^\circ) = \cos^2 15^\circ\cos 60^\circ = \dfrac14(1+\cos30^\circ),$$
$$g_m = \dfrac{2+\sqrt3}8 < \dfrac{2\sqrt3+\sqrt3}8 = f(30^\circ, 30^\circ).$$
The second bound gives the same value $g_m.$
Thus, in the issue constraints,
$$\boxed{\min \{\cos x\cos y\cos z\} = \dfrac{2-\sqrt3}8\approx 0.46656}$$
at 
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\in
\left\{
\begin{pmatrix}15^\circ\\15^\circ\\60^\circ\end{pmatrix},
\begin{pmatrix}15^\circ\\60^\circ\\15^\circ\end{pmatrix},
\begin{pmatrix}60^\circ\\15^\circ\\15^\circ\end{pmatrix}\right\}
$$
