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?

Also, we can make the following.
Let $b^2+c^2-a^2=x$, $a^2+c^2-b^2=y$ and $a^2+b^2-c^2=z$.
Thus, $x$, $y$ and $z$ are positives, $a=\sqrt{\frac{y+z}{2}},$ $b=\sqrt{\frac{x+z}{2}},$ $c=\sqrt{\frac{x+y}{2}}$ and  in the standard notation 
$$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{2S}{bc}=\sum_{cyc}\frac{\frac{1}{2}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}}{bc}=\sum_{cyc}\frac{\sqrt{xy+xz+yz}}{\sqrt{(x+z)(x+y)}}=$$
$$=\sqrt{\frac{(xy+xz+yz)\left(\sum\limits_{cyc}\sqrt{x+y}\right)^2}{\prod\limits_{cyc}(x+y)}}=\sqrt{\frac{(xy+xz+yz)\left(2(x+y+z)+2\sum\limits_{cyc}\sqrt{(x+y)(x+z)}\right)}{\prod\limits_{cyc}(x+y)}}>$$
$$>\sqrt{\frac{(xy+xz+yz)\left(2(x+y+z)+2(x+y+z)\right)}{\prod\limits_{cyc}(x+y)}}=2\sqrt{\frac{(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}}>2.$$
