Maximum value of $ (1- \tan A)(1- \tan B)(1- \tan C)$

Question

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.

Answer 1

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.