Proving $\pi(\frac1A+\frac1B+\frac1C)\ge(\sin\frac A2+\sin\frac B2+\sin\frac C2)(\frac 1{\sin\frac A2}+\frac 1{\sin\frac B2}+\frac 1{\sin\frac C2})$

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.)