In acute angled triangle, $\cos \frac{A}{2} > \sin \frac{A}{2}$
So, $\sin \frac{A}{2}.\cos \frac{A}{2} > \sin \frac{A}{2}.\sin \frac{A}{2}$
Hence, $2\sin \frac{A}{2}\cos \frac{A}{2} > 2{\sin ^2}\frac{A}{2}$
Thus, $\sin A > 1 - \cos A$
Likewise, $\sin B > 1 - \cos B$ & $\sin C > 1 - \cos C$
On addition, $\sin A + \sin B + \sin C > 3 - (\cos A + \cos B + \cos C)$ ........(!)
In any triangle, $\cos A + \cos B + \cos C \le \frac{3}{2}$*
$\therefore 2(\cos A + \cos B + \cos C) \le 3$
$ \Rightarrow \cos A + \cos B + \cos C \le 3 - (\cos A + \cos B + \cos C)$ ........(!!)
From (!) & (!!), $\sin A + \sin B + \sin C > \cos A + \cos B + \cos C$
*Please refer textbook or study material for the proof.