6. As shown in the figure, in isosceles $\triangle ABC$, $CD$ is the altitude on the base $AB$, $E$ is the midpoint of the leg $BC$, and $AE$ intersects $CD$ at $F$. Now, three routes are given:
(a) $A \rightarrow F \rightarrow C \rightarrow E \rightarrow B \rightarrow D \rightarrow A$;
(b) $A \rightarrow C \rightarrow E \rightarrow B \rightarrow D \rightarrow F \rightarrow A$;
(c) $A \rightarrow D \rightarrow B \rightarrow E \rightarrow F \rightarrow C \rightarrow A$;
Let their lengths be $L(a)$, $L(b)$, and $L(c)$, respectively. Among the following three inequalities: $L(a) < L(b)$, $L(a) < L(c)$, $L(b) < L(c)$, the number of inequalities that must hold is
(A) 0; (B) 1; (C) 2; (D) 3.
Answer ( )
6. (B)

From the given conditions, we know that $F$ is the centroid of $\triangle A B C$, so $C F=2 D F, A F=2 F E$. Since $C A=C B$, we also have
$$
\begin{array}{l}
L(c)=A B+B E+E F+F C+C A, \\
L(a)=A F+F C+C B+B A,
\end{array}
$$

Therefore, $L(c)-L(a)=B E+E F-A F=B E+E F-B F>0$, which means $L(c)>L(a)$.
Also, $L(b)=A C+C E+E B+B D+D F+F A$,
$$
\begin{aligned}
L(c)-L(b) & =(A B-B D)+(E F-F A)+(F C-D F)-C E \\
& =A D+D F-C E-E F,
\end{aligned}
$$

When $\triangle A B C$ is an equilateral triangle, $A D=C E, D F \doteq E F$, thus $L(c)=L(b)$. This indicates that $L(b)0,
$$

which means $L(a)>L(b)$, so the inequality $L(a)<L(b)$ does not always hold. Therefore, among the three inequalities, only one always holds.