<s>[INST] 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?
 [/INST] I have found a simpler solution.
Observing the graph of $y = \sin x$ and $y = \frac{2}{\pi}x$, $x\in[0,\frac{\pi}{2}]$.
We can see that $\sin x\ge\frac{2}{\pi}x$.
Let $\bigtriangleup ABC$ be a triangle such that each angle is less than $90^{\circ}$. So we have:
$A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),C\in(0,\frac{\pi}{2})$.
Because $\sin x>\frac{2}{\pi}x$, when $x\in(0,\frac{\pi}{2})$,
$$\sin A+\sin B+\sin C > \frac{2}{\pi}(A+B+C)=2$$
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