Let $n,k\geq 2$ be positive integers. For each $1\leq i\leq n$, let $I\_i$ be a nonempty subset of $\{1,2,\dots,k\}$. Let $P\_i=\sum\_{j\in I\_i}x\_j$, and let $P=P\_1\cdot P\_2\cdot\dots\cdot P\_n$. (For example, $P=x\_1(x\_1+x\_2)(x\_1+x\_3)$.) 


We want to maximize this expression subject to the constraints $x\_i\geq 0$ for all $i$, and $\sum\_{i=1}^k x\_i=1$. Let $A$ be the value of $P\_1$ at the maximum. Let $B$ be the value of $P\_1$ at the maximum if we instead maximize the expression $P'=P\_2\cdot P\_3\cdot\dots\cdot P\_n$, subject to the same constraints.



> 
> Is it true that $A\geq \frac{n-1}{n}B+\frac{1}{n}$?
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Equality can be obtained, e.g., for any value of $n$, when $P\_1=x\_1$ and $P\_i=x\_2$ for $i=2,\dots,n$ (so $A=1/n$ and $B=0$).


Another example: $n=2$, $P\_1=x\_1$ and $P\_2=x\_1+x\_2$. Then $A=1$ and $B$ can be anything in $[0,1]$, so the inequality always holds.



Maximize product of sums
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.


