This problem is trivial with barycentric coordinates. Let $ \triangle{ABC} $ be our reference triangle, so that $ A = (1, 0, 0) $ and $ B = (0, 1, 0) $ and $ C = (0, 0, 1). $ Also let $ a = BC, b = CA, $ and $ c = AB. $ It is well-known that $ G = (1 : 1 : 1) $ and $ I = (a : b : c) $ so we find that $ [AIG] = \frac{[ABC]}{3(a + b + c)}\begin{vmatrix} 1 & 0 & 0 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = \frac{[ABC]|b - c|}{3(a + b + c)}. $ Finding similar equations for $ [BIG] $ and $ [CIG] $ we find that $ [AIG] + [BIG] + [CIG] = \frac{[ABC](|b - c| + |c - a| + |a - b|)}{3(a + b + c)} = \frac{r(|b - c| + |c - a| + |a - b|)}{6}. $ Now assume WLOG that $ a \le b \le c. $ Since $ b - a \ge d $ and $ c - b \ge d $ we have that $ c - a \ge 2d $ which implies that $ \frac{r(|b - c| + |c - a| + |a - b|)}{6} \ge \frac{2dr}{3} $ as desired.

Equality iff $a-b=b-c$ where sides are named such that $a\ge b\ge c$