Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$. Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$.
I started from considering
$$\begin{align}\sin A+\sin B+\sin (180^o-A-B) &= \sin A+\sin B+\sin(A+B) 
\\&=\sin A+\sin B+\sin A\cos B + \cos A\sin B 
\\&=\sin A(1+\cos B)+\sin B(1+\cos A).\end{align}$$
How to proceed?

Note that since $\sin(x)$ is concave and $\left ( \frac{\pi}{2} , \frac{\pi}{2}, 0 \right ) \succ (A, B, C)$, then by Karamata's Inequality
$$\sin(A) + \sin(B) + \sin(C) > \sin \left ( \frac{\pi}{2} \right ) + \sin \left ( \frac{\pi}{2} \right ) + \sin(0) = 2.$$
