Showing that $ 1<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}$ I would like to show that:
$$ 1<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}$$
where $\alpha, \beta, \gamma$ are the angles of a triangle.
I know that the inequality $$ 1<\cos \alpha+\cos \beta+\cos \alpha $$
is a direct consequence of the identity $$ \cos \alpha+\cos \beta+\cos \alpha =1+\frac{r}{R}$$ 
with circumradius $R$ and inradius $r$.
So is there a similar expression for $$ \sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}?$$
 A: Rewrite the inequality as $\sin \frac{\alpha + \beta + \gamma}{2} < \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} + \sin \frac{\gamma}{2}$. Notice that $\sin (a+b) < \sin a + \sin b$ for $a, b, (a+b) \in (0, {\pi \over 2})$. Extend the same statement for three variables.
A: (Using $\alpha$, $\beta$, $\gamma$ to name the half-angles themselves ...)


$$\alpha + \beta + \gamma = 90^\circ \quad\implies\quad 1 \leq \sin \alpha + \sin\beta + \sin\gamma$$
  with equality when one of $\alpha$, $\beta$, $\gamma$ is $90^\circ$.

(By the Law of Sines, an angle $\theta$, inscribed in a circle of diameter $1$, subtends a chord of length $\sin\theta$.)
Note. More generally, for $\alpha+\beta+\gamma \leq 180^\circ$,
$$\sin\left(\alpha+\beta+\gamma\right) \leq \sin \alpha + \sin\beta + \sin\gamma$$
and even more generally, the relation clearly holds for an arbitrary number of angles.
A: In order to demonstrate the inequality above we may actually prove a more general inequality, which can come in handy another time:
For all natural $n > 1$ and $x_1, x_2, \ldots, x_n \in (0, \pi)$ we have:
\[ |\sin(x_1 + x_2 + ...+x_n)| < \sin x_1 + \sin x_2 + ... + \sin x_n.\]
This can be proved by induction:


*

*for $n = 2$ 
\[
\begin{split}
|\sin(x_1 + x_2)| &= |\sin x_1 \cos x_2 + \sin x_2 \cos x_1| \leq |\sin x_1 \cos x_2| + |\sin x_2 \cos x_1| \\
&= |\sin x_1| \cdot |\cos x_2| + |\sin x_2| \cdot |\cos x_1| < \sin x_1 + \cos x_2
\end{split}
\]
by the properties of absolute value and since $\cos x < 1$ for $x \in (0, \pi)$

*induction step
\[
\begin{split}
|\sin (x_1 + \ldots + x_{n+1})| &= |\sin(x_1 + \ldots + x_n) \cos x_{n+1} + \sin x_{n+1} \cos(x_1 + \ldots + x_n)| \\
&\leq |\sin(x_1 + \ldots + x_n)| \cdot |\cos x_{n+1}| + |\sin x_{n+1}| \cdot |\cos(x_1 + \ldots + x_n)| \\
&< |\sin(x_1 + \ldots + x_n)| + |\sin x_{n+1}| \\
&< \sin x_1 + \sin x_2 + \ldots + \sin x_{n+1}.
\end{split}
\] by the induction hypothesis and the fact $|\sin x| = \sin x$ for $x \in (0, \pi)$
When plugging in halves of the angles of a triangle we obtain the desired
\[ \sin(\dfrac{\alpha + \beta + \gamma}{2}) = 1 < \sin \dfrac{\alpha}{2} + \sin\dfrac{\beta}{2} + \sin\dfrac{\gamma}{2} .\]
A: Let $\alpha\geq\beta\geq\gamma$.
$\sin$ is a concave function on $\left(0^{\circ},90^{\circ}\right)$
and $\left(90^{\circ},0^{\circ},0^{\circ}\right)\succ\left(\frac{\alpha}{2},\frac{\beta}{2},\frac{\gamma}{2}\right)$.
Thus, by Karamata:
$$1=\sin90^{\circ}+\sin0^{\circ}+\sin0^{\circ}<\sin\frac{\alpha}{2}+\sin\frac{\beta}{2}+\sin\frac{\gamma}{2}.$$
Done!
