jmc

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 $$\begin{array}{rl}& a+p(a+b)+p(a+b)+q(b+c)+q(b+c)+q(b+c)+r(a+c)+r(a+c)+r(a+c)+r(a+c)\\ & \ge 10\sqrt[10]{a\cdot p^2(a+b{)}^2\cdot q^3(b+c{)}^3\cdot r^4(a+c{)}^4},\end{array}$$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 $\boxed{\frac{1}{64}}.$