<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] Put $x=$sin $(\frac {A}{2}) $ and similarly define $y $ and $z $. Then required inequality becomes
$2 (xy+yz+zx)\leq (x+y+z) $ which translates to $s^2\leq s+x^2+y^2+z^2$ where $s=x+y+z $. Since $\sum_{cyc}x^2\geq \frac {s^2}{3} $, it is enough to prove that $3s^2\leq 3s+s^2$. Sunce $s>0$ it is enough to prove that $s\leq \frac {3}{2} $. To prove this consider a $\triangle XYZ $ such that $2X=180^{\circ}-A $ and similarly. Then $s=$cos$X+$cos $Y+$cos $Z\leq 1+\frac {r}{R} $ wrt $\triangle XYZ $, so Euler's formula finishes the job.
Credit to https://artofproblemsolving.com/community/c6h1405767p7873009
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