6. In $\triangle A B C$, if $2 \cos A+\cos B$ $+\cos C=2$, then the equal quantitative relationship satisfied by the three sides $a, b, c$ is
$\begin{array}{l}6. \text { From } 2 \cos A+\cos B+\cos C=2 \text { we get } \\ \cos B+\cos C=2(1-\cos A), \\ 2 \cos \frac{B+C}{2} \cos \frac{B-C}{2}=4 \sin ^{2} \frac{A}{2}. \because \frac{A}{2}+\frac{B}{2} \\ +\frac{C}{2}=\frac{\pi}{2}, \therefore \cos \frac{B+C}{2}=\sin \frac{A}{2}, \sin \frac{B+C}{2} \\ =\cos \frac{A}{2}. \text { Therefore, } 2 \cos \frac{B-C}{2}=4 \sin \frac{A}{2}, \\ 2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}=4 \sin \frac{A}{2} \cos \frac{A}{2}, \text { that is } \\ \sin B+\sin C=2 \sin A. \therefore \text { By the Law of Sines, } b+c \\ =2 a .\end{array}$