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 $A\geq \frac{n-1}{n}B+\frac{1}{n}$?

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.

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Hans
    Aug 8, 2018 at 20:00
  • $\begingroup$ Doesn't $P=x_1(x_1+x_2)$ give $A=1$, $B\in[0,1]$ and hence violate this inequality? $\endgroup$ Aug 9, 2018 at 13:11
  • $\begingroup$ I guess all sets $I_i$ should be non-empty, yes? $\endgroup$ Aug 17, 2018 at 15:29
  • $\begingroup$ Which quantities are fixed in your optimization problem? $I_i$ ? $\endgroup$
    – Taro Tokyo
    Nov 7, 2018 at 12:25

1 Answer 1


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.

  • $\begingroup$ @nan The formula is just the second order Taylor formula, isn't it? If you want total rigor, you should care a bit about the error term, which I'm just ignoring, and spend some time explaining why $P$ is a locally Lipschitz function of $a$. As to "integration", we have the inequality $\frac{P'}{1-P}\ge \frac 1{n-1+a}$. What do you get when integrating this from $0$ to $1$? On the other hand I admit that it is rather sketchy, so feel free to ask more questions and I'll try to answer :-) $\endgroup$
    – fedja
    Nov 12, 2018 at 13:43
  • $\begingroup$ I'm not sure what we get when integrating $\frac{P'}{1-P}$. Erm... Then, I'm afraid, it's going to be a longer story than I thought it would. You get the increment of $-\log(1-P)$, so the result is $-\log(1-P(1))+\log(1-P(0))\ge\log\frac{n}{n-1}$, i.e., $1-P(1)\le \frac{n-1}n(1-P(0))$, which is equivalent to your inequality. Also we do not keep the sum of $a_k$ fixed in the argument. We just assume that it is $1$ at the original moment not to drag it along. $\endgroup$
    – fedja
    Nov 12, 2018 at 19:32
  • $\begingroup$ @nan We have the maximal value at $P$ and the $\tau$-perturbation can be used with both positive and negative $\tau$, so the first derivative has to be $0$. The inequality follows from $da_1\frac{dP_1}{P_1}\ge 2\frac 12(da_1)^2\frac{1-P_1}{P_1}$ (see the last two sentences in the argument) $\endgroup$
    – fedja
    Nov 12, 2018 at 21:02
  • $\begingroup$ @nan No, $P$ refers to $(P_1,\dots,P_m)$. Sorry for the change in the notation. Yes, by the "linear part" I mean this sum. The comparison is just the observation that the point $P+tdP=(P_1+tdP_1,\dots,P_m+tdP_m)$ shouldn't give a larger value to the RHS than $P+dP$, i.e., that when you move from $P$ in the direction of $dP$, you should reach the vertex of the parabola and stop there. Literally it is not true: you can guarantee only that the value of the RHS should not deviate from that maximum by more than $O((da_1)^3)$ (the cubic error), but that is enough. $\endgroup$
    – fedja
    Nov 12, 2018 at 22:41
  • $\begingroup$ The Taylor formula is just that $\log (X+dX)\approx\log X+\frac{dX}X-\frac 12\left(\frac{dX}{X}\right)^2$. $\endgroup$
    – fedja
    Nov 12, 2018 at 22:45

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