A geometry problem by Skanda Prasad

Using sine rule , we have:

$\begin{aligned} \frac c{\sin C} & = \frac a{\sin A} = \frac b{\sin B} \\ \implies \frac {\cot \frac A2 \cot \frac B2}{\sin C} & = \frac {\cot \frac B2 \cot \frac C2}{\sin A} = \frac {\cot \frac C2 \cot \frac A2}{\sin B} \\ \frac {\frac 1{\tan \frac A2 \tan \frac B2}}{\frac {2\tan \frac C2}{1+\tan^2 \frac C2}} & = \frac {\frac 1{\tan \frac B2 \tan \frac C2}}{\frac {2\tan \frac A2}{1+\tan^2 \frac A2}} = \frac {\frac 1{\tan \frac C2 \tan \frac A2}}{\frac {2\tan \frac B2}{1+\tan^2 \frac B2}} \\ \frac {1+\tan^2 \frac C2}{2\tan \frac A2 \tan \frac B2 \tan \frac C2} & = \frac {1+\tan^2 \frac A2}{2\tan \frac A2 \tan \frac B2 \tan \frac C2} = \frac {1+\tan^2 \frac B2}{2\tan \frac A2 \tan \frac B2 \tan \frac C2} \\ \implies 1 + \tan^2 \frac C2 & = 1 + \tan^2 \frac A2 = 1 + \tan^2 \frac B2 \\ \implies C & = A = B = 60^\circ \end{aligned}$

$\implies a=b=c = \cot^2 30^\circ = 3 \implies s = \dfrac 92$

$\implies \dfrac 1{s-a} + \dfrac 1{s-b} + \dfrac 1{s-c} = \dfrac 3{\frac 92 - 3} = \boxed{2}$