Question

If `<x_1<x_2<x_3<pi,` then prove that `sin((x^1+x_2+x_3)/3)<(sinx_1+sinx_2+sinx_3)/3` . Hence or otherwise prove that if `AdotBdotC` are angles of a triangle, then the maximum value of `sinA+sinB+sinC` is `(3sqrt(3))/2`

Answer 1

Draw the graph of y = sin x. `0 lt x lt pi` and take points `A(x_(1), sin x_(1)), B(x_(2), sin x_(2))` and `C(x_(3), sin x_(3)).`

Let the points A, B, C form a traingle.
y - coordinate of the centroid G is `(sin x_(1) + sin x_(2) + sin x_(3))/(3)`
and y - coordinate of the point f is `sin ((x_(1) + x_(2) + x_(3))/(3))` (as abscissa of F is `(x_(1) + x_(2) + x_(3))/(3)`)
From the diagram, `FD gt GD`.
Hence `sin ((x_(1) + x_(2) + x_(3))/(3)) gt (sin x_(1) + sin x_(2) + sin x_(3))/(3)`
If `A + B + C = pi`, then
`sin ((A + B + C)/(3)) gt (sin A + sin B + sin C)/(3)`
`rArr` `"sin" (pi)/(3) gt (sin A + sin B + sin C)/(3)`
`rArr` `(3sqrt(3))/(2) gt sin A + sin B + sin C`
`rArr` maximum value of `(sin A + sin B + sin C) = (3sqrt(3))/(2)`