How does the max of $\prod_i a_i$ work?

Question

Here are two succinct statements of the 'same' question:

Statement 1: Take $a>0$ and $S \subseteq \mathbb{R}^N; S=\{(x_1,\dots,x_N)| \frac{1}{N}\sum_i x_i = a; x_i>0\}$. Define a 'product function' $f:S\rightarrow \mathbb{R}^N; f(x_1,\dots,x_i)=\prod_ix_i$. There are many proofs that the max of $f$ over all of $S$ occurs at $\vec{v}_a=(a,a,\dots,a)$. But here is the 'extension':

Consider two points $c,d \in S$ with $||\vec{c}-\vec{v}_a||_1\leq||\vec{d}-\vec{v}_a||_1$ ('taxicab' $\ell_1$ norm). Why does $f(\vec{d})<f(\vec{c})<f(\vec{v}_a)$?

Statement 2: Take two sets of real numbers $C=\{c_1,\dots, c_N\}, D=\{d_1,\dots,d_N\}$ with $\sum_i c_i=\sum_i d_i = 0$ and $\sum_i|c_i|<\sum_i |d_i|$. Take some $a>0$ great enough so that $a+x_i>0$ for any $x_i\in C \cup D$.

Why is $\prod_i (a+d_i) < \prod_i(a+c_i)$?

Apparently it is a known result, but after a day of grief on my own I haven't been able to make any useful progress. My most promising approaches seemed to be:

• Fiddling with the AM-GM inequality
• Halve $C,D$ about their means (or centers?) to get more sets. Then try and somehow use the fact that for $a>b$, $(a-b)(a+b)<a^2$. Iterate recursively until you have the desired product.
• Show that $f$'s derivative is largest in the $(1,1,\dots, 1)$ direction, $\vec{c},\vec{d}$. Then show that in $S$ you are 'closest' to this direction the smaller $\ell_1$-distance you are from $\vec{v}_a$. Note that $S$ is a positive hyperplane, but it's shifted so it's not a subspace.
• Not-so-promising was expanding the polynomial product into a sum of $2^N$ $N^{\mathrm{th}}$-degree terms.

Answer 1

Your statements are false. Consider for example $\vec{c}=(2,100,498)$ and $\vec{d}=(25,25,550)$. Both $\vec{c}$ and $\vec{d}$ have mean $a=200$. And $||\vec{c}-\vec{a}||_1=596<700=||\vec{d}-\vec{a}||_1$, but $f(\vec{c})=99600<343750=f(\vec{d})$.

You probably forgot some additional conditions.

Answer 2

$\ln x$ is an increasing function. therefore $\arg \max\prod\limits_{i=1}^{x_n}x_i = \arg \max \ln \prod\limits_{i=1}^N x = \arg\max\sum\limits_{i=1}^n\ln x_i$. Once you've noticed it, it should be easier.