Distance or time?

Let $t_A$ and $t_B$ be the times it took for Brothers $A, B$ to cover distance $d$ , while $w, r$ are their speeds of walking and running. Then

$t_A=\dfrac{\frac{d}{2}}{w}+\dfrac{\frac{d}{2}}{r}$
$t_B=\dfrac{2d}{w+r}$

The ratio works out to

$\dfrac{t_A}{t_B}=\dfrac{(w+r)^2}{4wr}$

so that $t_A>t_B$ *, meaning that Brother B gets there first.

*Note: Given any positive $x, y$ ,

$\frac{1}{2}(x+y)>\sqrt{xy}$

Assume fot the sake of simplicity that they both run with velocity $v$ , and walk with vilocity $\frac{v}{2}$ . Brother A: Assume that brother A runs in $t_{1}$ time, and walks in $t_{2}$ time. Then, $\frac{s}{2}=\frac{v}{2} \cdot t_{1}$ $\frac{s}{2} = v \cdot t_{2}$ => $t_{1} = \frac{s}{v}$ $t_{2} = \frac{s}{2v}$ Total time= $\frac{3s}{2v}$

Brother B: $S_{1} = \frac{v}{2} \cdot \frac{t}{2}$ $S_{2} = v \cdot \frac{t}{2}$ => Total $S = \frac{3}{4} \cdot v \cdot t$ => $t= \frac{4s}{3v}$ ===> Brother B is faster.