This has to be well-known, no? a) Let the centroid be $G$. The distance of $G$ to $BC$ is one-third of $h_a$, the distance from $A$ to $BC$. Hence by defining similarly $h_b, h_c$ we have to show that $h_a+h_b+h_c \geq 9r$, with $r$ the inradius of $ABC$. Now if $\Delta$ is the area of $ABC$ then we have $ah_a = bh_b = ch_c = 2\Delta$, so $(a,b,c),(h_a,h_b,h_c)$ are oppositely sorted. Hence by Rearrangement Inequality, \[(a+b+c)(h_a+h_b+h_c) \geq 3(ah_a+bh_b+ch_c) = 6\Delta\]\[ \implies h_a+h_b+h_c \geq 3\frac{2\Delta}{a+b+c} = 9r,\] as desired. b) Draw $ABC$ with $B$ to the left of $C$. For a line $l_a$ parallel to $BC$ intersecting $AB,AC$ at $B',C'$ respectively, and a variable point $P$ on $l_a$, we see that: 1. The distance $d(P,BC)$ from $P$ to $BC$ is constant; 2. $d(P,AB)$ is linear in $PB'$, and $d(P,AC)$ is linear in $PC'$. Hence moving $P$ to the left of $B'$ or to the right of $C'$ increases both distances, and hence the sum of distances from $P$ to the sides. Thus we only need to consider points in or on the boundary of $ABC$. 3. Furthermore, since the sum of distances from $P$ to the sides is linear on $B'C'$, it attains its minimum at either $B'$ or $C'$. Say it attains its minimum at $B'$; then repeating the above argument with $l_c = AB$, we see that the sum of distances to the sides takes its minimum at either $A$ or $B$. A similar argument holds if the minimum is attained at $C'$. Thus the minimum sum of distances is $\min(h_a,h_b,h_c)$. If $BC$ is the unique longest side in $ABC$ then the unique point minimizing the sum of distances to the sides is $A$. If $BC = CA > AB$ then all points on segment $AB$ minimize the sum. If $ABC$ is equilateral then all points inside or on the boundary of $ABC$ minimize the sum.