Question about sines of angles in an acute triangle 
Let $\triangle ABC$ be a triangle such that each angle is less than $ 90^\circ $.
  I want to prove that $\sin A + \sin B + \sin C > 2$.

Here is what I have done:
Since $A+B+C=180^{\circ}$ and $0 < A,B,C < 90^\circ$, at least two of $A,B,C$ are in the range 45 < x < 90, without loss of generality, let these angles be $A$ and $B$.
$\sin A + \sin B + \sin C = \sin A + \sin B + \sin(180^\circ-A-B) = \sin A + \sin B + \sin(A+B)$
Since $45^\circ < A,B < 90^\circ$ it follows that $2^{0.5} < \sin A + \sin B < 2.$ Am I near the answer?

i am able to simplify  $$
\sin A + \sin B + \sin (A + B)  = \sin A + \sin B + \sin A \cos B + \sin B  \cos A \\ = (1+\cos B)\sin A + (1 + \cos A)\sin B 
\\ = 4\cos^2 B/2\sin A/2 \cos A/2 + 4\cos^2 A/2 \sin B/2 \cos B/2
\\ = 4\cos B/2 \cos A/2(\sin A/2 \cos B/2 + \sin B/2 \cos A/2)
\\ = 4\cos B/2 \cos A/2\sin (A/2 + B/2)  = 4\cos A/2 \cos B/2 \cos C/2
$$
4now, using the fact that $A/2 < 45^\circ, B/2 < 45^\circ$ and $C/2 < 45^\circ,$ i can only conclude $$\sin A + \sin B + \sin C > \sqrt 2.$$
i am going to try to improve the bound. introduce $0 <\alpha, \beta < 45^\circ$ so that $A/2 = 45^\circ - \alpha, B/2 = 45^\circ - \beta, C/2 = \alpha + \beta.$ in terms of these new variables, 
$$ 4\cos A/2 \cos B/2 \cos C/2 = 2(\cos \alpha + \sin \alpha)(\cos \beta + \sin \beta)\cos(\alpha + \beta)
\\ = 2(\cos(\alpha + \beta) + \sin(\alpha + \beta))\cos(\alpha + \beta)
\\ = 1 + \cos(2\alpha + 2\beta) + \sin(2\alpha + 2\beta) = 1 + \sqrt 2 \sin(2\alpha + 2\beta + 45^\circ)$$
since $0 < 2\alpha + 2 \beta < 90^\circ,$ we get the desired bound 
$$2 < 1 + \sqrt 2\sin(2\alpha + 2 \beta) < 1 + \sqrt 2 $$
