If $A+B+C = \pi$, show that $\sum_{\text{cyclic}} \tan A \tan B \neq 0$ 
If $A+B+C = \pi$, and $A,B,C > 0$ show that $$\sum_{\text{cyclic}} \tan A \tan B \neq 0$$
without using the fact that if $A+B+C = \pi$, then $\tan A + \tan B + \tan C = \tan A \tan B \tan C$
My Work

I tried to show it as follows:
$$\tan(A+B+C) = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \ tan C}{1-\sum_{\text{cyclic}} \tan A \tan B}$$
For $$A+B+C = \pi$$
$$0 = \dfrac{\tan A + \tan B + \tan C - \tan A \tan B \ tan C}{1-\sum_{\text{cyclic}} \tan A \tan B}$$
I can't proceed from here. 

First
$$
\tan C=\tan(\pi-A-B)=-\tan(A+B).
$$
Hence
$$
0=\tan A\tan B+\tan A\tan C+\tan C\tan B=\tan A\tan B+\tan C(\tan A+\tan B) \\=\tan A\tan B-\frac{(\tan A+\tan B)^2}{1-\tan A\tan B}
$$
Thus
$$
(\tan A+\tan B)^2=\tan A\tan B(1-\tan A\tan B)
$$
or
$$
(\tan A)^2+(\tan B)^2+\tan A\tan B=-(\tan A)^2(\tan B)^2.
$$
Contradiction, since the right-hand side is negative, while the left hand side is positive.
