Five. (20 points) In a mountain bike race held in a city, two cyclists, A and B, start from point A to point B at the same time. Cyclist A runs $\frac{1}{3}$ of the time at speeds $v_{1}, v_{2}, v_{3}$ respectively; Cyclist B runs $\frac{1}{3}$ of the distance at speeds $v_{1}, v_{2}, v_{3}$ respectively. Who will reach point B first, A or B?
Let the distance between $A$ and $B$ be $S$. The times taken by car A and car B to complete the journey are $t_{1}$ and $t_{2}$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{t_{1}}{3}\left(v_{1}+v_{2}+v_{3}\right)=S, \\
\frac{S}{3}\left(\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}\right)=t_{2} .
\end{array}\right.
$$

From (1), we get $t_{1}=\frac{3 S}{v_{1}+v_{2}+v_{3}}$.
From (2), we get $t_{2}=\frac{S\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right)}{3 v_{1} v_{2} v_{3}}$.
$$
\begin{array}{l}
t_{1}-t_{2} \\
=\frac{S\left[9 v_{1} v_{2} v_{3}-\left(v_{1}+v_{2}+v_{3}\right)\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right)\right]}{3 v_{1} v_{2} v_{3}\left(v_{1}+v_{2}+v_{3}\right)} \\
=\frac{-S\left[v_{1}\left(v_{2}-v_{3}\right)^{2}+v_{2}\left(v_{1}-v_{3}\right)^{2}+v_{3}\left(v_{1}-v_{2}\right)^{2}\right]}{3 v_{1} v_{2} v_{3}\left(v_{1}+v_{2}+v_{3}\right)}
\end{array}
$$
$\leqslant 0$.
From this, we can conclude that when $v_{1}, v_{2}, v_{3}$ are not all equal, $t_{1}<t_{2}$, meaning car A arrives at $B$ first. When $v_{1}=v_{2}=v_{3}$, $t_{1}=t_{2}$, meaning both car A and car B arrive at $B$ simultaneously.