Example 6 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, while Cyclist B runs $\frac{1}{3}$ of the distance at speeds $v_{1}, v_{2}, v_{3}$ respectively. Which cyclist, A or B, will arrive at B first?
(1993. Sichuan Province Junior High School Mathematics League Competition)
Solution: Let the distance between the two places be $s$, and the time taken by A and B to complete the entire journey be $t_{1}$ and $t_{2}$, respectively. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{1}{3} t_{1}\left(v_{1}+v_{2}+v_{3}\right)=s, \\
\frac{s}{3}\left(\frac{1}{v_{1}}+\frac{1}{v_{2}}+\frac{1}{v_{3}}\right)=t_{2}
\end{array}\right. \\
\therefore t_{1}=\frac{3 s}{v_{1}+v_{2}+v_{3}}, \\
\quad t_{2}=\frac{s\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right)}{3 v_{1} v_{2} v_{3}} .
\end{array}
$$

And $t_{1}-t_{2}$
$$
\begin{array}{l}
=\frac{s\left[9 v_{1} v_{2} v_{3}-\left(v_{1}+v_{2}+v_{3}\right)\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right)\right]}{3 v_{1} v_{2} v_{3}\left(v_{1}+v_{2}+v_{3}\right)} \\
=\frac{-s\left[v_{1}\left(v_{2}-v_{3}\right)^{2}+v_{2}\left(v_{1}-v_{3}\right)^{2}+v_{3}\left(v_{1}-v_{2}\right)^{2}\right]}{3 v_{1} v_{2} v_{3}\left(v_{1}+v_{2}+v_{3}\right)}<0,
\end{array}
$$

Therefore, $t_{1}<t_{2}$.