How does the max of $\prod_i a_i$ work? Here are two succinct statements of the 'same' question:

Statement 1:
Take $a&gt;0$ and $S \subseteq \mathbb{R}^N; S=\{(x_1,\dots,x_N)| \frac{1}{N}\sum_i x_i = a; x_i&gt;0\}$. 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 $f(\vec{d})&lt;f(\vec{c})&lt;f(\vec{v}_a)$?


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 $\sum_i|c_i|&lt;\sum_i |d_i|$. Take some $a&gt;0$ great enough so that $a+x_i&gt;0$ for any $x_i\in C \cup D$. 


  Why is $\prod_i (a+d_i) &lt; \prod_i(a+c_i)$?


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 $a&gt;b$, $(a-b)(a+b)&lt;a^2$. 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.
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 $||\vec{c}-\vec{a}||_1=596&lt;700=||\vec{d}-\vec{a}||_1$, but $f(\vec{c})=99600&lt;343750=f(\vec{d})$.

You probably forgot some additional conditions.
$\ln x$ 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.