Maximum of $\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B$ in triangle What is the maximum value of $$\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B,$$ where $A,B,C$ are angles in a triangle?
We can rewrite as $$-\sin A\sin B\sin(A+B)+\sin B\sin(A+B)\cos A+\sin(A+B)\sin A\cos B$$
Expanding, this becomes
$$-\sin^2 A\sin B\cos B-\sin^2B\sin A\cos A+2\sin B\sin A\cos B\cos A+\cos^2A\sin^2B+\sin^2A\cos^2B$$
 A: I will need the fact that, if $A,B,C$ are the angles of a triangle, then
$$ \cos A + \cos B + \cos C \le \tfrac32 \tag{$\ast$} $$
with equality iff $A=B=C=\frac\pi3$.  With this result in hand, your problem can be solved thus:
\begin{align*}
&\cos A\sin B\sin C + \sin A\cos B\sin C + \sin A\sin B\cos C \\
&= \cos A\cos B\cos C - \cos(A+B+C) \\
&= \cos A\cos B\cos C + 1 \\
&\le \left(\frac{\cos A+\cos B+\cos C}{3}\right)^3 + 1
  &&\text{(AM/GM)} \\
&\le \tfrac98
  &&\text{(by ($\ast$))}
\end{align*}
with, again, equality iff $A=B=C=\frac\pi3$.
To prove ($\ast$), first note that since $A+B+C=\pi$ and all three are positive, at most one of them is greater than $\frac\pi2$.  Wlog, $A,B\in[0,\frac\pi2]$.  On this interval, $\cos$ is concave, so by Jensen's inequality,
\begin{align*}
\cos A + \cos B + \cos C
&\le 2\cos(\tfrac{A+B}{2}) + \cos C \\
&= 2\cos(\tfrac{A+B}{2}) - \cos(A+B) \\
&= 2\cos(\tfrac{A+B}{2}) - 2\cos^2(\tfrac{A+B}{2}) + 1 \\
&= \tfrac32 - 2\left(\cos(\tfrac{A+B}{2})-\tfrac12\right)^2 \\
&\le \tfrac32
\end{align*}
with equality iff $A=B=C=\frac\pi3$.
A: Let $\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B=y$
$$\implies2y=\cos C[\cos(A-B)-\cos(A+B)]+2\sin C[\sin B\cos A+\sin A\cos B]$$
Now $\sin B\cos A+\sin A\cos B=\sin(A+B)=\sin(\pi-C)=\sin C$ and $\cos(A+B)=\cos(\pi-C)=-\cos C$
$$\implies y=\cos C[\cos(A-B)+\cos C]+2\sin C[\sin C]$$
On rearrangement, $$\cos^2C-\cos C\cos(A-B)+2y-2=0$$ which is a Quadratic Equation in $\cos C$
As $C$ is real, $\implies\cos C$ is real, the discriminant,  $\cos^2(A-B)-4(2y-2)$ must be non-negative
$\implies 4(2y-2)\le\cos^2(A-B)$ which is $\le1$
$\implies y\le\dfrac98$  the equality occurs if $\cos(A-B)=\pm1$
But as $0<A,B<\pi,$
$(i)\cos(A-B)\ne-1$  and consequently $(ii)\cos(A-B)=1\implies A-B=0$
In that case, $\cos C=\dfrac{\cos(A-B)}2=\dfrac12\implies C=?$
