I found it quite easy to show that AP = a+b+c, because if b and c are the mass of points B and C respectively, P is the sum of their masses (b+c). Then segement AP has a mass of a+b+c because it includes point A with a mass of a, and point P with mass b+c. Unless that is completely off.
I'm having a difficult time trying to determine how AI=b+c.
Keep in mind, point I should be 2/3 of the distance on AP.
You say that if $b,c$ are masses at $B,C$ then the mass of $P$ is $b+c$. This is not necessarily true because $P$ will not necessarily be the "balance", the center of mass, between $B$ and $C$. For example, if $b=c$, then their center of mass is in the middle between them, which is not where $P$ is $(AP$ is an angle bisector, not a median!$)$.
To solve this right, you need to put masses in vertices $A,B,C$ in a clever way. The idea is that for example the balance between $B$ and $C$ $($the center of mass of just these two$)$ should be exactly at $P$. If that is true, this immediately tells you that the center of mass of all three points lies somewhere on the bisector $AP$, because you can find it by first finding that of $B,C$, and then adding $A$. Once you know that, it follows that the angle bisectors must intersect at one point $I$, because the center of mass of all three vertices has to lie somewhere, and it has to lie on every bisector.
Now, since you want the center of mass of just $B,C$ to lie at $P$, the lengths $BP$, $CP$ must be inversely proportional to the masses $B$, $C$ $($the law of the lever!$)$. But the lengths $BP$, $CP$ are themselves proportional to $AB$, $AC$, because $AP$ is the bisector $($this is a basic property of angle bisectors in a triangle$)$. So actually the masses at $B$, $C$ can be inversely proportional to lengths $AB$, $AC$ respectively. For example, if $AB$ is $3$ times longer than $AC$, the mass at $B$ should be $3$ times smaller than $C$. This may give you an idea of how to select the masses for all three vertices.