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!
Best Answer
$\sin\left(\alpha\right) \geq 2\alpha/\pi\,,\quad\sin\left(\beta\right) \geq 2\beta/\pi\,,\quad\sin\left(\gamma\right) \geq 2\gamma/\pi$.
$$ \sin\left(\alpha\right) + \sin\left(\beta\right) + \sin\left(\gamma\right) \geq 2\,{\alpha + \beta + \gamma \over \pi} = 2 $$
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 $$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \sin\left(\alpha\right) + \sin\left(\beta\right) + \sin\left(\gamma\right) \color{#000000}{\ >\ } 2\quad} \\ \\ \hline \end{array} $$