Yes, it does follow. For $i=1,2$, let $P_i, Q_i, R_i$ be the vertices of $T_i$ opposite the sides of length $a_i, b_i, c_i$, respectively. We first check the case where $a_1 = a_2$ (or $b_1 = b_2$ or $c_1 = c_2$, by the same argument after relabeling). Imagine $T_2$ as being drawn with the base $Q_2R_2$ horizontal and the point $P_2$ above the line $Q_2R_2$. We may then position $T_1$ so that $Q_1 = Q_2$, $R_1 = R_2$, and $P_1$ lies above the line $Q_1R_1 = Q_2R_2$. Then $P_1$ also lies inside the region bounded by the circles through $P_2$ centered at $Q_2$ and $R_2$. Since $\angle Q_2$ and $\angle R_2$ are acute, the part of this region above the line $Q_2R_2$ lies within $T_2$. In particular, the distance from $P_1$ to the line $Q_2R_2$ is less than or equal to the distance from $P_2$ to the line $Q_2R_2$; hence $A_1 \leq A_2$.