4. In an acute $\triangle A B C$, find the minimum value of $3 \tan B \tan C+2 \tan A \tan C+\tan A \tan B$ is $\qquad$
Answer: $6+2 \sqrt{3}+2 \sqrt{2}+2 \sqrt{6}$
Solution: From the identity $\tan B+\tan C+\tan A=\tan C \tan A \tan B$, we know that $\frac{1}{\tan B \tan C}+\frac{1}{\tan A \tan C}+\frac{1}{\tan A \tan B}=1$. Therefore, by the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{c}
(3 \tan B \tan C+2 \tan A \tan C+\tan A \tan B)\left(\frac{1}{\tan B \tan C}+\frac{1}{\tan A \tan C}+\frac{1}{\tan A \tan B}\right) \\
\geq(\sqrt{3}+\sqrt{2}+1)^{2}=6+2 \sqrt{3}+2 \sqrt{2}+2 \sqrt{6}
\end{array}
$$