If we apply AM-GM to one instance of $ pa, $ two instances of $ q(a + b), $ three instances of $ r(b + c), $ and four instances of $ s(a + c), $ then we get \begin{align*} &a + p(a + b) + p(a + b) + q(b + c) + q(b + c) + q(b + c) + r(a + c) + r(a + c) + r(a + c) + r(a + c) \\ &\ge 10 \sqrt[10]{a \cdot p^2 (a + b)^2 \cdot q^3 (b + c)^3 \cdot r^4 (a + c)^4}, \end{align*}where $ p, $ $ q, $ and $ r $ are constants to be decided. In particular, we want these constants so that \[a + p(a + b) + p(a + b) + q(b + c) + q(b + c) + q(b + c) + r(a + c) + r(a + c) + r(a + c) + r(a + c)\]is a multiple of $ a + b + c $. This expression simplifies to \[(1 + 2p + 4r) a + (2p + 3q) b + (3q + 4r) c.\]Thus, we want $ 1 + 2p + 4r = 2p + 3q $ and $ 2p + 3q = 3q + 4r $. Then $ 2p = 4r, $ so $ p = 2r $. Then \[1 + 8r = 3q + 4r,\]so $ q = \frac{4r + 1}{3} $. For the equality case, \[a = p(a + b) = q(b + c) = r(a + c).\]Then $ a = pa + pb, $ so $ b = \frac{1 - p}{p} \cdot a $. Also, $ a = ra + rc, $ so $ c = \frac{1 - r}{r} \cdot a $. Substituting into $ a = q(b + c), $ we get \[a = q \left( \frac{1 - p}{p} \cdot a + \frac{1 - r}{r} \cdot a \right).\]Substituting $ p = 2r $ and $ q = \frac{4r + 1}{3}, $ we get \[a = \frac{4r + 1}{3} \left( \frac{1 - 2r}{2r} \cdot a + \frac{1 - r}{4} \cdot a \right).\]Then \[1 = \frac{4r + 1}{3} \left( \frac{1 - 2r}{2r} + \frac{1 - r}{r} \right).\]From this equation, \[6r = (4r + 1)((1 - 2r) + 2(1 - r)),\]which simplifies to $ 16r^2 - 2r - 3 = 0 $. This factors as $ (2r - 1)(8r + 3) = 0 $. Since $ r $ is positive, $ r = \frac{1}{2} $. Then $ p = 1 $ and $ q = 1, $ and AM-GM gives us \[\frac{a + (a + b) + (a + b) + (b + c) + (b + c) + (b + c) + \frac{a + c}{2} + \frac{a + c}{2} + \frac{a + c}{2} + \frac{a + c}{2}}{10} \ge \sqrt[10]{\frac{a (a + b)^2 (b + c)^3 (a + c)^4}{16}}.\]Hence, \[\sqrt[10]{\frac{a (a + b)^2 (b + c)^3 (a + c)^4}{16}} \le \frac{5(a + b + c)}{10} = \frac{1}{2}.\]Then \[\frac{a (a + b)^2 (b + c)^3 (a + c)^4}{16} \le \frac{1}{2^{10}} = \frac{1}{1024},\]so \[a (a + b)^2 (b + c)^3 (a + c)^4 \le \frac{16}{1024} = \frac{1}{64}.\]Equality occurs when \[a = a + b = b + c = \frac{a + c}{2}.\]Along with the condition $ a + b + c = 1, $ we can solve to get $ a = \frac{1}{2}, $ $ b = 0, $ and $ c = \frac{1}{2} $. Hence, the maximum value is $ \boxed{\frac{1}{64}} $.