6. Given $2 n$ positive real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}$ and $0 \leqslant \lambda \leqslant 2$, let
$$
M=\sum_{i=1}^{n} \sqrt{a_{i}{ }^{2}+b_{i}{ }^{2}-\lambda a_{i} b_{i}}, N=\sum_{i=1}^{n} \sqrt{a_{i=1}^{2}+b_{i}^{2}-\lambda a_{i+1} b_{i}} \text {, }
$$

where, $a_{n+1}=a_{1}$. Then the relationship between $M$ and $N$ is
A. $M \geqslant N$
B. $M \leqslant N$
C. depends on $\lambda$
D. completely uncertain
6. B. Construct $\angle A O B$, with the included angle $\theta=\arccos \frac{\lambda}{2} \in[0, \pi]$.

As shown in the figure, take $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ and $B_{1}, B_{2}, \cdots, B_{n}$ on the rays $O A$ and $O B$ respectively, such that $O A_{i}=a_{i}, O B_{i}=b_{i}$. By the cosine rule, we get $N \geqslant M$, that is,
$$
A_{2} B_{1}+A_{3} B_{2}+\cdots+A_{1} B_{n} \geqslant A_{1} B_{1}+A_{2} B_{2}+\cdots+A_{n} B_{n} \text {. }
$$

Let $A_{1} B_{n}$ intersect $A_{2} B_{1}, A_{3} B_{2}, \cdots, A_{n} B_{n-1}$ at points $X_{1}, X_{2}, \cdots, X_{n-1}$ respectively. Then,
$$
\begin{array}{c}
A_{2} B_{1}+A_{3} B_{2}+\cdots+A_{1} B_{n}=\left(A_{1} X_{1}+B_{1} X_{1}\right)+\left(A_{2} X_{1}+X_{1} X_{2}+\right. \\
\left.X_{2} B_{2}\right)+\cdots+\left(A_{n} X_{n-1}+X_{n-1} B_{n}\right) \geqslant A_{1} B_{1}+A_{2} B_{2}+\cdots+A_{n} B_{n}
\end{array}
$$

Thus, the conclusion holds.