<s>[INST] 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 $\cos x\cos y\cos z$.
 [/INST] Note that $x\geqslant y\geqslant z\geqslant15°$ and $x + y + z = 90°$ implies $2y + z\leqslant90°$ and $15°\leqslant z\leqslant30°$, then\begin{align*}
\cos x\cos y\cos z &= \frac12(\cos(x + y) + \cos(x - y))\cos z\\
&= \frac12(\cos(90° - z) + \cos(90° - 2y - z))\cos z\\
&= \frac12(\sin z + \sin(2y + z))\cos z \geqslant \frac12(\sin z + \sin 3z)\cos z\\
&= \sin 2z\cos z\cdot\cos z = 2\sin z\cos^3 z.
\end{align*}
Now define $f(z) = \sin z\cos^3 z$. For $15°\leqslant z\leqslant30°$,\begin{align*}
f'(z) &= \cos^4 z - 3\sin^2 z\cos^2 z = (\cos^2 z - 3\sin^2 z)\cos^2 z\\
&= (1 - 4\sin^2 z)\cos^2 z \geqslant 0,
\end{align*}
thus denoting $z_0 = 15°$, then\begin{align*}
&\mathrel{\phantom{=}}{} \cos x\cos y\cos z \geqslant 2\sin z\cos^3 z \geqslant 2\sin z_0\cos^3 z_0\\
&= \sin 2z_0 \cos^2 z_0 = \sin 2z_0 \cdot \frac12(\cos 2z_0 + 1)\\
&= \frac12 \cdot \frac12\left( \frac{\sqrt3}2 + 1 \right) = \frac{2 + \sqrt3}8.
\end{align*}
Because for $(x, y, z) = (60°, 15°, 15°)$, $\cos x\cos y\cos z = \dfrac{2 + \sqrt3}8$, then the minimum is indeed $\dfrac{2 + \sqrt3}8$.
</s>