Maximum value of $\sin A+\sin B+\sin C$ without calculus or Lagrange Multipliers

Question

Let $A, B, C$ be a the angles of a triangle. Find the maximum value of $\sin A+\sin B+\sin C$ without calculus or Lagrange Multipliers.
My attempt: If we fix $A$, then $B+C$ is also fixed. We have $\sin B+ \sin C=2\sin\frac{B+C}{2}\cos\frac{B-C}{2}$. The maximum value of $\cos\frac{B-C}{2}$ is $1$ and it is attained when $B=C$. What should I do next?

Answer 1

(1) $\sin A+\sin B+\sin C$ is the maximum when the area of the quadrilateral $PQRS$ is the maximum.

(2) With $A$ fixed, the area of $PQRS$ is the maximum when the area of $\triangle QRS$ is the maximum, i.e. when $R$ is the farthest point on the arc $QS$ from the chord $QS$. So, $R$ is the midpoint of arc $QS$. So, $B=C$.

(3) The area remains unchanged if $A$ and $B$ are interchanged. $PQRS$ is then an isosceles trapezium inscribed in the semi-circle.

(4) Reflect the trapezium in $PS$ to form a hexagon inscribed in the unit circle.

(5) The area of $PQRS$ is equal to the area of $QRST$. If it is maximized, $QRST$ should also be an isosceles trapezium. SO, $\angle RSP=\angle QOP$. Note that $\angle RSP=\angle QPS=\angle PQT$. Therefore, $\triangle OPQ$ is equilateral.

$\sin A+\sin B+\sin C$ is the maximum when $A=B=C=\dfrac{\pi}{3}$.