5. Let $x_{i} \in \mathbf{R}, x_{i} \geqslant 0(i=1,2,3,4,5), \sum_{i=1}^{j} x_{i}=1$, then $\max \left\{x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{1}\right.$, $\left.x_{4}+x_{5}\right\}$ has the minimum value equal to
A. $\frac{1}{4}$
B. $\frac{1}{3}$
C. $\frac{1}{6}$
D. $\frac{1}{8}$
5. $\mathrm{B}$ Hint: $\max \left\{x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}, x_{4}+x_{5}\right\} \geqslant \max \left\{x_{1}+x_{2}, x_{3}+x_{4}, x_{4}+x_{3}\right\} \geqslant$ $\frac{1}{3}\left(\sum_{i=1}^{5} x_{i}+x_{4}\right) \geqslant \frac{1}{3}$. Taking $x_{1}=x_{3}=x_{5}=\frac{1}{3}, x_{2}=x_{4}=0$, then $\max \left\{x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}, x_{4}+\right.$ $\left.x_{5}\right\}=\frac{1}{3}$, so the answer is B.