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?
 A: 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
$$\sin(A) + \sin(B) + \sin(C) > \sin \left ( \frac{\pi}{2} \right ) + \sin \left ( \frac{\pi}{2} \right ) + \sin(0) = 2.$$
A: Let's maximize the continuous function $f(A,B)=\sin A + \sin B + \sin(A+B)$ subject to the constraints $0\le A\le{\pi\over2}$, $0\le B\le{\pi\over2}$, and ${\pi\over2}\le A+B\le\pi$, that is, on the triangular region with vertices $({\pi\over2},{\pi\over2})$, $(0,{\pi\over2})$, and $({\pi\over2},0)$. This region is the closure of the region in the $AB$-plane of points $(A,B)$ containing possible angles $A$ and $B$ of a non-degenerate acute triangle $\triangle ABC$.
If we set the partial derivatives to zero, we get $\cos A+\cos(A+B)=\cos B+\cos(A+B)=0$, whence $\cos A=\cos B$ and thence $\cos (2A)=-\cos(A)$. This easily yields a unique local maximum $f({\pi\over3},{\pi\over3})$ and no local minima in the interior of the domain.
The domain of $f$ is closed, so $f$ attains a minimum, and because there is no local minimum in the interior of the domain, $f$ attains its minimum at a boundary point. At every boundary point, one of the angles is $\pi\over2$, so the minimum value of $f$ is f($x_m$,${\pi\over2}-x_m)$ for some $x_m$ with $0\le x_m\le{\pi\over2}$. Replacing $x_m$ with ${\pi\over2}-x_m$ leaves the set of angles unchanged, so without loss of generality, assume that $x_m<{\pi\over2}-x_m$.
Because $\sin$ is concave down on the interval $[0,{\pi\over2}]$, $\sin({x_m-\epsilon})+\sin({{\pi\over2}-x_m-\epsilon}) < \sin(x_m)+\sin({{\pi\over2}+x_m})$ for any $0<\epsilon\le x_m$. Because $f$ was minimized with $x_m$, no such $\epsilon$ can exist for which $x_m-\epsilon\ge0$, and therefore $x_m=0$. This means that $f(0,{\pi\over2})=\sin0+\sin{\pi\over2}+\sin0=2$ is the minimum value of $f$ on its domain. This domain is a superset of the domain of points that correspond to non-degenerate acute triangles, so that domain is also minimized by the value 2.
A: 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 
$$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{2S}{bc}=\sum_{cyc}\frac{\frac{1}{2}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}}{bc}=\sum_{cyc}\frac{\sqrt{xy+xz+yz}}{\sqrt{(x+z)(x+y)}}=$$
$$=\sqrt{\frac{(xy+xz+yz)\left(\sum\limits_{cyc}\sqrt{x+y}\right)^2}{\prod\limits_{cyc}(x+y)}}=\sqrt{\frac{(xy+xz+yz)\left(2(x+y+z)+2\sum\limits_{cyc}\sqrt{(x+y)(x+z)}\right)}{\prod\limits_{cyc}(x+y)}}>$$
$$>\sqrt{\frac{(xy+xz+yz)\left(2(x+y+z)+2(x+y+z)\right)}{\prod\limits_{cyc}(x+y)}}=2\sqrt{\frac{(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}}>2.$$
