If A+B+C=π Prove that in triangle ABC, sin `A+sin B+Cle(3sqrt(3))/(2)`

Explanation

In `Delta ABC`, we have `0 lt A,B,C lt pi`.
So, consider the graph of function `y=sin x ` for `x in (0,pi)`.
The graph of function is concave downward.
Now, consider three points on the graph `P(A, sin A),Q(B,sin B) and R(C, sin C)`.

Join these points to get triangle PQR.
Now, centroid of the triangle PQR is
`G((A+B+C)/(3),(sin A+sin B+sin C)/(3))`
Through point G, draw a line parallel to y-axis meeting graph at point D and x-axis at point E.
The abscissa of point D is `(A+B+C)/(3)`
So, coordinates of point D are `((A+B+C)/(3),sin ((A+B+C)/(3)))`
Clearly, `DE ge GE`,
`therefore sin((A+B+C)/(3)) ge (sin A+sin B+sinC)/(3)`
`rArr sin (pi)/(3) ge (sin A sin B +sin C)/(3)`
`rArr sin A +sin B +sin C le (3sqrt(3))/(2)`.