Cars A and B travel the same distance. Car A travels half that distance at $u$ miles per hour and half at $v$ miles per hour. Car B travels half the time at $u$ miles per hour and half at $v$ miles per hour. The average speed of Car A is $x$ miles per hour and that of Car B is $y$ miles per hour. Then we always have 
$\textbf{(A)}\ x \leq y\qquad \textbf{(B)}\ x \geq y \qquad \textbf{(C)}\ x=y \qquad \textbf{(D)}\ xy$
Let $t$ be the total number of time in hours that Car B took to drive the distance.  This means that for have the time Car B traveled $\tfrac{ut}{2}$ miles and for half the time Car B traveled $\tfrac{vt}{2}$ miles.  That means the total distance traveled is $\tfrac{ut+vt}{2}$ miles, so
\[y = \frac{\frac{ut+vt}{2}}{t}\]
\[y = \frac{u+v}{2}\]
Because both cars traveled the same distance, for half the distance Car A took $\tfrac{ut+vt}{4u}$ hours and for half the distance Car A took $\tfrac{ut+vt}{4v}$ hours.  That means
\[x = \frac{\frac{ut+vt}{2}}{\frac{ut+vt}{4u} + \tfrac{ut+vt}{4v}}\]
\[x = \frac{2}{\frac{1}{u} + \frac{1}{v}}\]
In order to compare $x$ and $y$, we need to compare the values that $x$ and $y$ are equal to.
\[\frac{2}{\frac{1}{u} + \frac{1}{v}} \bigcirc \frac{u+v}{2}\]
Since the denominators of both numbers are positive, cross-multiplying won’t change the comparison sign.
\[4 \bigcirc 2 + \frac{u}{v} + \frac{v}{u}\]
\[2 \bigcirc  \frac{u}{v} + \frac{v}{u}\]
Because $uv$ is positive, the comparison sign does not need to be changed either.
\[2uv \bigcirc u^2 + v^2\]
\[0 \bigcirc (u-v)^2\]
By the [Trivial Inequality](https://artofproblemsolving.com/wiki/index.php/Trivial_Inequality), $0 \le (u-v)^2$.  All of the steps are reversible, so $\boxed{\textbf{(A)}\ x \leq y}$.  This can be confirmed by testing values of $u$ and $v$.