Proving $\pi(\frac1A+\frac1B+\frac1C)\ge(\sin\frac A2+\sin\frac B2+\sin\frac C2)(\frac 1{\sin\frac A2}+\frac 1{\sin\frac B2}+\frac 1{\sin\frac C2})$ Let $\Delta ABC$, prove that
$$\pi\left(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right)\ge \left(\sin{\dfrac{A}{2}}+\sin{\dfrac{B}{2}}+\sin{\dfrac{C}{2}} \right) \left(\dfrac{1}{\sin{\dfrac{A}{2}}}+\dfrac{1}{\sin{\dfrac{B}{2}}}+\dfrac{1}{\sin{\dfrac{C}{2}}}\right)$$
 A: Maybe this will help you,
$$\sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \le \dfrac {1}{8}$$(With equality iff $A=B=C$)
Well here's a proof of it:
We have 
$$\sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2}$$
$$=\dfrac{1}{2}[\cos \dfrac{A-B}{2}-\cos \dfrac {A+B}{2}]\sin \dfrac{C}{2}$$
$$\le \dfrac{1}{2}[1-\sin \dfrac{C}{2}]\sin \dfrac{C}{2}$$
$$=\dfrac{1}{2}[\sin \dfrac{C}{2}-\sin^2 \dfrac{C}{2}]$$
$$=\dfrac{1}{2}[ \dfrac{1}{4} -(\dfrac{1}{2}- \sin \dfrac{C}{2})^2] \le \dfrac{1}{8}$$
$$A+B+C= \pi$$
Maximum value$(ABC)=\dfrac{\pi^3}{27}$(From AM-GM inequality)
Maximum value of $(AB+BC+CA) =\dfrac{\pi^2}{9}$
$$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\dfrac{AB+BC+CA}{ABC}$$
We have 
$$\dfrac{1}{ABC} \ge \dfrac{27}{\pi^3}$$ 
$$(AB+BC+CA) \ge \dfrac{\pi^2}{3}$$
(You can multiply inequalities, when they are both positive)
We get, $$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C} \ge \dfrac{\pi^2}{3}.\dfrac{27}{\pi^3}$$
$$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge 9 \dfrac{1}{\pi}$$
(Multiplying $\pi$)
$$ \pi. (\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}) \ge 9$$
You can carry on from here. (Manipulation of inequalities, nothing much.)
