Let $A$, $B$, and $C$ be the angles of a triangle, with angle $C$ as the smallest of them. Show that
(i) $\sin \left(\frac{C}{2}\right) \leq \frac{1}{2}$
(ii) Hence, or otherwise, show that $$\sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)<\frac{1}{4}$$
By applying Jensen's inequality, I am getting
$$\sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)\leq\frac{1}{8}$$
iff $A=B=C=\pi/3$.
If the smallest angle $C\gt60^\circ$, it means that $A+ B<120^\circ$, which menas that either $A$ or $B$ must be smaller than $60^\circ$ i.e. smaller than $C$ which is a contradiction. So it must be that $C\le60^\circ$ and therefore:
$$\sin\frac C2\le\sin 30^\circ=\frac 12\tag{1}$$.
Now we know $A+B=180^\circ - C$
$$2\sin\frac A2\sin \frac B2=\cos\frac{A-B}2-\cos\frac{A+B}2=\cos\frac{A-B}2-\cos\frac {180^\circ - C}2=$$
$$\cos\frac{A-B}2-\sin\frac C2\tag{2}$$
Consider the maximum value the last expression in (2) can get. Obviously the first item should be as big as possible and the second one as small as possible. The first tiem can have a maximum value of 1 (in case $A=B$). For all other values of $A,B$, the first term is smaller. The second term is some positive number, definitively between 0 and $\frac 12$.
So the maximum value the expression (2) can get is actually smaller than 1 (the biggest value is achieved for $B=C$ and infinitesimally small value of angle $C$). For arbitrary values of $A,B,C$ it is now obvius that: $$2\sin\frac A2\sin \frac B2\lt 1$$
$$\sin\frac A2\sin \frac B2\lt \frac 12\tag{3}$$
By combining (1) and (3) we get:
$$\sin\frac A2\sin\frac B2\sin\frac C2\lt\frac14$$