19. Person A and Person B start from the same place and follow the same route to the same destination at the same time. Person A walks half of the time at speed $m$ and the other half at speed $n$. Person B walks half of the distance at speed $m$ and the other half at speed $n$. If $m \neq n$, who will arrive at the destination first? What if $m=n$?
19. Let the distance from the starting point to the designated location be $S$, and the time taken by person A and person B to complete the journey be $t_{1}, t_{2}$, respectively. Then: $\frac{t_{1}}{2} m+\frac{t_{1}}{2} n=S, \frac{S}{2 m}+\frac{S}{2 n}=t_{2} \quad$ We can derive: $t_{1}=\frac{2 S}{m+n}, t_{2}=\frac{S(m+n)}{2 m n}$. Therefore, $t_{1}-t_{2}=\frac{2 S}{m+n}-\frac{S(m+n)}{2 m n}=\frac{S\left[4 m n-(m+n)^{2}\right]}{2(m+n) m n}=-\frac{S(m-n)^{2}}{2 m n(m+n)}$. Since $S, m, n$ are all positive numbers, and $m \neq n$, it follows that $t_{1}-t_{2}<0$ which means: $t_{1}<t_{2}$. Thus, person A arrives at the designated location first.