$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ 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

Let $a=\frac{\frac{1}{y}+\frac{1}{z}}{2},$ $b=\frac{\frac{1}{x}+\frac{1}{z}}{2}$ amd $c=\frac{\frac{1}{x}+\frac{1}{y}}{2},$ where $x$, $y$ and $z$ are positives,
Thus, we need to prove that:
$$\sum_{cyc}z\sqrt{x+y}\geq2\sum_{cyc}\frac{xy}{\sqrt{x+y}}$$ or
$$\sum_{cyc}\frac{y(z-x)-x(y-z)}{\sqrt{x+y}}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{z}{\sqrt{y+z}}-\frac{z}{\sqrt{x+z}}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2z\sqrt{x+y}\geq0.$$
