smc

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 $\frac{ut}{2}$ miles and for half the time Car B traveled $\frac{vt}{2}$ miles. That means the total distance traveled is $\frac{ut+vt}{2}$ 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 $\frac{ut+vt}{4u}$ hours and for half the distance Car A took $\frac{ut+vt}{4v}$ hours. That means $$x=\frac{\frac{ut+vt}{2}}{\frac{ut+vt}{4u}+\frac{ut+vt}{4v}}$$ $$x=\frac{2}{\frac{1}{u}+\frac{1}{v}}$$ In order to compare $x$ and $y$, we need to compare the values that $x$ and $y$ are equal to. $$\frac{2}{\frac{1}{u}+\frac{1}{v}}◯\frac{u+v}{2}$$ Since the denominators of both numbers are positive, cross-multiplying won't change the comparison sign. $$4◯2+\frac{u}{v}+\frac{v}{u}$$ $$2◯\frac{u}{v}+\frac{v}{u}$$ Because $uv$ is positive, the comparison sign does not need to be changed either. $$2uv◯u^2+v^2$$ $$0◯(u-v{)}^2$$ By the Trivial Inequality, $0\le (u-v{)}^2$. All of the steps are reversible, so $\boxed{x\le y}$. This can be confirmed by testing values of $u$ and $v$.