6. In the acute triangle $\triangle A B C$, let the sides opposite to $\angle A, \angle B, \angle C$ be $a, b, c$ respectively. If $\frac{b}{a}+\frac{a}{b}=4 \cos C$, then the minimum value of $\frac{1}{\tan A}+\frac{1}{\tan B}$ is $\qquad$
6. $\frac{2}{\sqrt{3}}$.

It is known that, $\cos C=\frac{1}{4}\left(\frac{b}{a}+\frac{a}{b}\right) \geqslant \frac{1}{2}$.
Then $\sin C \leqslant \frac{\sqrt{3}}{2}$.
Also, $a^{2}+b^{2}=4 a b \cos C \Rightarrow a^{2}+b^{2}=2 c^{2}$.
Therefore, $\frac{1}{\tan A}+\frac{1}{\tan B}=\frac{\cos B \cdot \sin A+\sin B \cdot \cos A}{\sin A \cdot \sin B}$ $=\frac{\sin (A+B) \cdot \sin C}{\sin A \cdot \sin B \cdot \sin C}=\frac{\sin ^{2} C}{\sin A \cdot \sin B \cdot \sin C}$
$=\frac{c^{2}}{a b \sin C}=\frac{a^{2}+b^{2}}{2 a b \sin C} \geqslant \frac{2 a b}{2 a b \sin C}$ $=\frac{1}{\sin C} \geqslant \frac{2}{\sqrt{3}}$.
The equality holds if and only if $\angle A=\angle B=\angle C$.