Maximum value of $ (1- \tan A)(1- \tan B)(1- \tan C)$ Assume that we have a triangle with angles A, B, C. so $A+B+C = \pi$.
What is the maximum value of $(1- \tan A)(1- \tan B)(1- \tan C)$?
My try:
I know that given $A+B+C = \pi$ we have that
$\tan A + \tan B + \tan C = \tan A \tan B \tan C$ and $\tan C = - \tan(A+B)$
Using the equalities above, I came to the result
$(1- \tan A)(1- \tan B)(1- \tan C) = 1 - 2(\tan A+ \tan B +\tan C )+(\tan A \tan B + \tan A \tan C +\tan B \tan C)$
and this is where I'm stuck, using the second equality I didn't get anything helpful.
I'm trying to solve this without using differentiation. any hints would be appreciated.

The maximum value is infinity. Take
$$A = \frac{\pi}{2} - \epsilon, B = \frac{\pi}{3} + \epsilon, C = \frac{\pi}{6}$$
for $\epsilon > 0$ arbitrarily small.
The minimum value is also infinity. Take
$$A = \frac{\pi}{2} + \epsilon, B = \frac{\pi}{3} - \epsilon, C = \frac{\pi}{6}$$
for $\epsilon > 0$ arbitrarily small.
