Yes, that is true. Here is a sketch of the argument. It needs some polishing in places but, I hope, it makes clear what is going on here.

We will show that if $P_1,P_2,\dots, P_m$ are of the above form and maximize $F=\sum_k a_k\log P_k$ (so they are viewed as functions of $a=(a_1,\dots,a_m)\in (0,+\infty)^m$), then $\frac{\partial P_1}{\partial a_1}\ge (1-P_1)(\sum_k a_k)^{-1}$ (to be precise, we are talking about the right derivative here). The desired result can be obtained from here by an appropriate integration.

Since everything is homogeneous in $a$, we can assume without loss of generality that $\sum_k a_k=1$. Let $da_1$ be a small (infinitesimal) positive increment of $a_1$. Let $dP_i$ be some admissible increments of $P_i$ ($P_i$ are running over some convex polyhedron, so locally the set of admissible increments is an infinitesimal cone in general). Then we have the following increment of $F$ (up to third order terms):
$$
dF-da_1\log P_1\approx
\sum_k a_k\frac {dP_k}{P_k}+
da_1\frac {dP_1}{P_1}-\frac 12\sum_k a_k\left(\frac {dP_k}{P_k}\right)^2\,.
$$
The true increment $dP$ should essentially maximize the RHS. Notice that if you consider the true maximizer $dP=(dP_1,\dots,dP_m)$, then you can compare it with $tdP$ for $t$ close to $1$ and obtain that the linear in $dP$ part of the RHS should be twice as large as the maximum itself (all I'm saying here is that $ut-\frac 12vt^2$ attains its maximal value $\frac {u^2}{2v}$ at $t=u/v$
where the linear part $ut$ equals $\frac{u^2}{v}$). Since we must have
$$
\sum_k a_k\frac {dP_k}{P_k}\le 0
$$
because $P$ is the point of maximum and the admissible domain is convex, we conclude that at the true $dP$, we must have $da_1dP_1$ at least as large as twice the maximum of the RHS.

Now it remains to estimate that maximum using some particular perturbation. The simplest one will be $x_i\mapsto (1+\tau)x_i$ for $x_i$ participating in $P_1$ and $x_j\mapsto \left(1-\tau\frac{P_1}{1-P_1}\right)x_j$ for $x_j$ not participating in $P_1$. Note that this perturbation is admissible for both small positive and small negative $\tau$ unless one of the variables is $1$, which means that all $P_k=1$, in which case there is nothing to prove. Thus for $dP$ corresponding to this perturbation, we have
$$
\sum_k a_k\frac {dP_k}{P_k}=0\,.
$$
We also have $\frac{dP_1}{P_1}=\tau$ and $-\frac{P_1}{1-P_1}\tau\le\frac{dP_k}{P_k}\le\tau$ for all $k$. Recalling that for a mean zero real function $f$ on $[0,1]$ squeezed between $-u$ and $v$, we have $\int_0^1 f^2\le uv$, we conclude that for this perturbation
$$
\frac 12\sum_k a_k\left(\frac{dP_k}{P_k}\right)^2\le \frac 12\frac{P_1}{1-P_1}\tau^2
$$
whence we can take $\tau=da_1\frac{1-P_1}{P_1}$ and see that the maximum of the RHS is at least $\frac 12 (da_1)^2\frac {1-P_1}{P_1}$. Since $da_1\frac{dP_1}{P_1}$ for the true maximizing increment should be at least twice that large, we get $dP_1\ge (1-P_1)da_1$, as desired.