The problem is easy to state:
Prove that $$\sin{\alpha}+\sin{\beta}+\sin{\gamma}>2$$ Where $\alpha$, $\beta$ and $\gamma$ are angles from an acute-angled triangle.
I only managed to turn it into: $$ a+b+c>4R $$ Where $a$, $b$ and $c$ are the sides of the triangle and $R$ is the radius of the circumcircle.
I was looking for a cool proof rather than a bunch of calculations! Thank you in advance for your help!
$\sin\left(\alpha\right) \geq 2\alpha/\pi\,,\quad\sin\left(\beta\right) \geq 2\beta/\pi\,,\quad\sin\left(\gamma\right) \geq 2\gamma/\pi$.
$$ \sin\left(\alpha\right) + \sin\left(\beta\right) + \sin\left(\gamma\right) \geq 2\,{\alpha + \beta + \gamma \over \pi} = 2 $$
The equal $\left(~=~\right)$ sign is excluded since it requires $\alpha = \beta = \gamma = 0$ or $\alpha = \beta = \gamma = \pi/2$ which do not satisfy the problem conditions.
Then $$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \sin\left(\alpha\right) + \sin\left(\beta\right) + \sin\left(\gamma\right) \color{#000000}{\ >\ } 2\quad} \\ \\ \hline \end{array} $$
A more geometric solution: $\triangle ABC$ is acute-angled so the circumcenter $O$ lies inside it. Now consider the midpoints $M$, $N$, $P$ and it's clear that circumcenter $O$ lies inside one of the quadrilaterals $BPNC$, $BMNA$, $APMC$, for example $O \in BPNC$
We have the inequality $$BO + OC < BP + PN + NC$$ Note that $PN$ it's a middle line so $PN = \frac{a}{2}$ and $$R + R < \frac{c}{2} + \frac{a}{2} + \frac{b}{2}$$ Using the law of sines we finally get $$2<\sin A + \sin B + \sin C$$
By using law of sines, $\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$, one can substitute the values for $\sin(A)$ and $\sin(B)$ in terms of $\sin(C)$ from the law of sines in the expression $\sin(A)+\sin(B)+\sin(C)$ and arrive at $\sin(C)(\frac{a+b+c}{c})$; where $a,b,c$ are the sides and $A,B,C$ are the opposite angles to them respectively.
Now using triangle law of inequality, $\sin(C)(\frac{a+b+c}{c})>\sin(C)*2$ . Assuming that $C$ is one of the angles that is at least $60$ degrees and atmost 90 degrees (without any loss of generality), we arrive at the answer since $\sin(60)=\frac{\sqrt{3}}{2}$.
Once you got a+b+c>4R there is nothing more. we have, a/sinα=b/sinβ=c/sinγ=2R from here we have
sum(sinα)=(a+b+c)/2R
Substituting this will lead to answer