$3 \cdot 27$ Cars $A$ and $B$ travel the same distance. Car $A$ travels half the distance at $u$ miles per hour and the other half at $v$ miles per hour. Car $B$ travels half the time at $u$ miles per hour and the other half at $v$ miles per hour. The average speed of car $A$ is $x$ miles per hour, and the average speed of car $B$ is $y$ miles per hour. Then we always have
(A) $x \leqslant y$.
(B) $x \geqslant y$.
(C) $x=y$.
(D) $xy$.
(24th American High School Mathematics Examination, 1973)
[Solution] Let $s$ represent the distance, and $t$ be the time car $A$ travels. According to the problem, we have
$$
t=\frac{S}{2 u}+\frac{S}{2 v},
$$

The average speed of car $A$ is $\frac{S}{t}=\frac{S}{\frac{S}{2 u}+\frac{S}{2 v}}=\frac{2 u v}{u+v}=x$.
Let $S_{1}, S_{2}$ represent the distances car $B$ travels at speeds $u, v$, respectively. Therefore, if the travel time of car $B$ is $T$, then
$$
\frac{S_{1}}{\frac{T}{2}}=u, \quad \frac{S_{2}}{\frac{T}{2}}=v .
$$

Adding the two equations gives $\frac{S}{\frac{T}{2}}=\frac{S_{1}+S_{2}}{\frac{T}{2}}=u+v$.
The average speed of car $B$ is $\quad \frac{S}{T}=\frac{u+v}{2}=y$.
For positive numbers $u$ and $v$, by $4 u v \leqslant(u+v)^{2}$, thus $x \leqslant y$. Therefore, the answer is $(A)$.