<s>[INST] $\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ Let $A,B,C$ be the three angles of a triangle. Prove that:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin\dfrac{C}{2}+\sin\dfrac{A}{2}\sin\dfrac{C}{2})$
Here all I did:
$p=\dfrac{a+b+c}{2}$
$\sin \dfrac{A}{2} = \sqrt { \dfrac {(p-b)(p-c)}{bc}} $
$\Rightarrow \sin \dfrac{A}{2} \sin \dfrac{B}{2} = \dfrac{(p-c)}{c} \sqrt { \dfrac {(p-a)(p-b)}{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {2p-a-b}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {c}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2}\le \dfrac{(p-c)}{\sqrt{ba}}$
So we need to prove that : $\sum \dfrac{(p-c)}{\sqrt{ba}} \le \sum\sqrt { \dfrac {(p-b)(p-c)}{bc}} $. But I still have no idea (I'm not sure that's true either). I hope to get help from everyone. Thanks a lot
 [/INST] Here's an alternative that take advantage of convexity of sine in $(0,\pi)$
$$
\begin{align}
\left(\sin{\frac{A}{2}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}\right)\cdot 2\cdot\sin{\frac{A/2+B/2+C/2}{3}}&\geq\frac{2}{3}\left(\sin{\frac{A}{2}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}\right)^{2}\\
&\geq 2\left(\sin{\frac{A}{2}\cdot\sin{\frac{B}{2}}+\sin{\frac{B}{2}}\cdot\sin{\frac{C}{2}}+\sin{\frac{C}{2}}}\cdot\sin{\frac{A}{2}}\right)
\end{align}
$$
Recall that $2\sin{\frac{A/2+B/2+C/2}{3}}=2\sin{\frac{\pi}{6}}=1$
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