Example 5 In the border desert area, patrol vehicles travel 200 kilometers per day, and each patrol vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their mission, and then return along the original route to the base. To allow three of them to patrol as far as possible (and then return together), vehicles Jia and Yi travel to a certain point $B$ along the way, leaving only enough gasoline for their return to the base, and leaving the extra gasoline for the other three vehicles to use. How far can the other three vehicles travel at most?
(1905, Hebei Province Junior High School Mathematics Joint Competition)
Analysis: The key is where point $B$ is most appropriate. Suppose car A and car B travel for $x$ days to reach point $B$. After reaching point $B$, considering the fuel capacity of the patrol car, we should have
$$
\frac{2(14-2 x)}{3}+(14-x) \leqslant 14 \text {. }
$$

Thus, $x \geqslant 4$.
For the other three cars, the fuel consumption for each car throughout the process is
$$
\frac{2(14-2 x)}{3}+14 \text {. }
$$

Therefore, the distance these three cars leave from base $A$ is
$$
\begin{aligned}
f(x) & =\frac{\frac{2(14-2 x)}{3}+14}{2} \times 200 \\
& =\frac{100}{3}(70-4 x) .
\end{aligned}
$$

This is a decreasing function of $x$.
Thus, $f(x)_{\max }=1800$ (km).
Note: Establishing indeterminate equations and inequalities based on the equal distance relationship can solve the problem of the maximum travel distance.
(2) Inequality relationship