<s>[INST] In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$ 
In $\triangle ABC$, if 
  $$\cos A \cos B \cos C=\frac{1}{3}$$ 
  then can we find value of 
  $$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$

Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition. 
 [/INST] The answer to the question is no. The maximum value of $\cos A \cos B \cos C$, where $A$, $B$, and $C$ are the angles of a triangle in the plane, is $\frac{1}{8}$, so there is no plane triangle for which $\cos A \cos B \cos C=\frac{1}{3}$.
The product $\cos A \cos B \cos C$ equals $\frac{1}{8}$ for an equilateral triangle, and the fact that this is a maximum follows from the fact that for any acute* triangle $\triangle ABC$, the product is greater for the “more equilateral” triangle with angles $\frac{A+B}{2}$, $\frac{A+B}{2}$, and $C$, because $$\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)\cos C-\cos A\cos B\cos C= \frac{1-\cos(A+B)}{2}\cdot\cos C>0.$$
*We can assume the triangle is acute (and $\cos C>0$), because otherwise the product $\cos A \cos B \cos C \le 0$ (only one angle in a given triangle can be non-acute, so only one of the cosines can be non-positive) and $\triangle ABC$ can’t possibly be one for which $\cos A \cos B \cos C$ is a maximum.
This might be an interesting question for triangles on a surface of negative curvature, however.
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