If xi>0, 1≤i≤n, and x1+x2+x3…+xn=π, then the greatest value of sinx1+sinx2+sinx3+...+sinxn=
n
π
nsinπn
0
We have sinx1+sinx2+sinx3+….+sinxnn ≤sinx1+x2+x3+….+xnn since sin is concave in 0,π ≤sinπn ∵x1+x2+x3+….+xn=π⇒sinx1+sinx2+sinx3+….+sinxn≤nsinπn