### Instruction: Among all triangles $ABC,$ find the maximum value of $\sin A + \sin B \sin C.$.### Response: This is a very interesting problem, and it involves some trigonometric identities and inequalities.
One thing I notice is that $\sin A + \sin B \sin C$ is symmetric in $A, B, C,$ so it does not depend on the order or the labels of the angles.
Another thing I notice is that $\sin A + \sin B \sin C \leq \sin A + 1,$ because $\sin B \sin C \leq 1$ for any angles $B$ and $C.$
So, to maximize $\sin A + \sin B \sin C,$ I need to maximize $\sin A$ and make $\sin B \sin C$ as close to $1$ as possible.
The maximum value of $\sin A$ is $1,$ and it occurs when $A = 90^\circ.$
If $A = 90^\circ,$ then $B$ and $C$ are complementary angles, and $\sin B \sin C = \cos B \cos C.$
The maximum value of $\cos B \cos C$ is also $1,$ and it occurs when $B = C = 45^\circ.$
So, the maximum value of $\sin A + \sin B \sin C$ is attained when $ABC$ is a right isosceles triangle, and it is equal to $1 + 1 = 2.$