Product / GM of numbers, with fixed mean, increase as numbers get closer to mean. I am trying to prove a statement which goes like this.
Let $a_i$ and $b_i$ be positive real numbers where $i = 1,2,3,\ldots,n$; where $n$ is a positive integer greater than or equal to $2$, such that,
$$0\lt a_1\le a_2 \le \ldots\le a_n \ \ and \ \ 0\lt b_1\le b_2 \le\ldots\le b_n \tag{i}$$
and, 
$$\sum^n_{i=1} a_i = \sum^n_{j=1} b_j \tag{ii}$$
If $\exists$ $k \in \Bbb Z$ such that $1 \le k\le n-1$ and, 
$$a_k\le b_1 \le b_2 \le \ldots \le b_n \le a_{k+1} \tag{iii}$$
then,
$$\prod^n_{i=1} a_i \le \prod^n_{j=1} b_j \tag{iv}$$
Is there anyway to prove this?
Comments: Basically we are trying to prove if the product of $n$ numbers whose mean is constant increases as all the numbers come closer (in the number line) to the mean than the closest number (out of the $n$ number in the previous step) to mean in the previous step.
It is a tweak of the famous $GM \le AM$ inequality.
 A: If we only 'move' two elements of the sequence $(a_k)_k$, letting $b_i:=a_i$ except for $i=j,k$ ($j<k$) when $b_j:=a_j+\varepsilon$ and $b_k:=a_k-\varepsilon$, then we have
$$b_jb_k=(a_j+\varepsilon)(a_k-\varepsilon) = a_ja_k+\varepsilon\,(a_k-a_j\, -\varepsilon) \ > \ a_ja_{k}$$
using that $a_k>a_j+\varepsilon$. (Actually, we can assume that $\varepsilon\le \displaystyle\frac{a_k-a_j}2 $, provided $a_j<a_k$.)
By induction on $n$ we might be able to prove that the given $(b_k)$ sequence can be obtained from the given $(a_k)$ sequence using repeatedly the 'move two elements' method.
A: I got a proof to my statement. It uses Karamata's inequality.
Karamata's Inequality (courtesy Wikipedia)
Statement of the Karamata's inequality goes like this.
Let $I$ be an interval of the real line and let $f$ denote a real-valued, convex function defined on $I$. If $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ are numbers in $I$ such that $(x_1, \ldots, x_n)$ majorizes $(y_1, \ldots, y_n)$, then
$$f(x_1)+\cdots+f(x_n) \ge f(y_1)+\cdots+f(y_n) \tag{1}$$
Here majorization means that
$$x_1+\cdots+x_n = y_1+\cdots+y_n \tag{2}$$
and, after relabeling the numbers $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$, respectively, in decreasing order, i.e.,
$$x_1 \ge x_2 \ge \cdots \ge x_n \ and\ y_1\ge y_2\ge\cdots\ge y_n, \tag{3}$$
we have
$$x_1+\cdots+x_i \ge y_1+\cdots+y_i \ \  \forall \ \ i \in \{1,2, \ldots, n-1\} \tag{4}$$

Proof
Here in this question take $(x_1, x_2, \ldots, x_n) = (a_n, a_{n-1}, \ldots, a_1)$ and $(y_1, y_2, \ldots, y_n) = (b_n, b_{n-1}, \ldots, b_1)$
(I) Conditions for majorization
(2) From (ii) $$\sum^n_{i=1} a_i = \sum^n_{j=1} b_j \implies x_1+\cdots+x_n = y_1+\cdots+y_n \tag{a}$$
(3) From (i)
$$a_n \ge a_{n-1} \ge \cdots \ge a_1 \ and\ b_n\ge b_{n-1}\ge\cdots\ge b_1 $$
$$\implies x_1 \ge x_2 \ge \cdots \ge x_n \ and\ y_1\ge y_2\ge\cdots\ge y_n$$
(4)
Let $X_i = x_1+\cdots+x_i$ and $Y_i = y_1+\cdots+y_i \ \  \forall \ \ i \in \{1,2, \ldots, n\}$.
From (i) and (iii)
$$a_{j} \ge b_{j} \ \  \forall \ \ j \in \{k+1, k+2, \ldots, n\}$$
$$\implies a_n + a_{n-1} + \cdots + a_{j} \ge b_n + b_{n-1} + \cdots + b_{j} \ \  \forall \ \ j \in \{k+1, k+2, \ldots, n\}$$
$$\implies x_1+\cdots+x_i \ge y_1+\cdots+y_i \ \  \forall \ \ i \in \{1,2, \ldots, n-k\} \tag{b}$$
From (i) and (iii),
$$a_{j} \le b_{j} \ \  \forall \ \ j \in \{1, 2, \ldots, k\}$$
$$\implies X_i-X_{i-1} \le Y_i - Y_{i-1} \ \  \forall \ \ i \in \{n-k+1, n-k+2, \ldots, n\} \tag{c}$$
This means that $Y_i$ increases faster than $X_i$ in the range $\{n-k+1, n-k+2, \ldots, n\} $
But from (a) and (b) know that $Y_n = X_n$ and  $Y_{n-k} \le X_{n-k}$. This along with (c) means that,
$$\implies X_i \ge Y_i \ \  \forall \ \ j \in \{n-k+1, n-k+2, \ldots, n-1\} \tag{d}$$
If the above statement was not true then there exist $l \in \{n-k+1, n-k+2, \ldots, n-1\}$ such that, $X_i \lt Y_i$. But then from (c) $Y_n$ can never be equal to $X_n$, as given in (a). This is a contradiction. Thus the above statement (d) is true.
Therefore, from (d) $$\implies x_1+\cdots+x_i \ge y_1+\cdots+y_i \ \  \forall \ \ i \in \{n-k+1,n-k+2, \ldots, n-1\} \tag{e}$$
From (b) and (e),
$$\implies x_1+\cdots+x_i \ge y_1+\cdots+y_i \ \  \forall \ \ i \in \{1, 2, \ldots, n-1 \}$$
Thus $(x_1, \ldots, x_n)$ majorizes $(y_1, \ldots, y_n)$.
Now take $f(x) = -ln(x) \ \ \forall \ \ x \in \mathbb {R^+}$. $f(x)$ is a convex function in $\mathbb {R^+}$because,
(II) Convexity of $f(x)$
(A) Derivative of $f(x)$,
$$f'(x) = - \frac1x \lt 0 \ \ \forall \ \ x \in \mathbb {R^+} \tag{f}$$
(B) Second derivative of $f(x)$,
$$f''(x) = \frac1{x^2} \gt 0 \ \ \forall \ \ x \in \mathbb {R^+}$$
(III) Using Karamata's Inequality
From (1)
$$f(x_1)+\cdots+f(x_n) \ge f(y_1)+\cdots+f(y_n) $$
$$\implies f(a_n)+\cdots+f(a_1) \ge f(b_n)+\cdots+f(y_1) $$
$$\implies -ln(a_n)-\cdots-ln(a_1) \ge -ln(b_n)-\cdots-ln(b_1) $$
$$\implies ln(a_n)+\cdots+ln(a_1) \le ln(b_n)+\cdots+ln(b_1) $$
$$\implies ln(\prod^n_{i=1} a_i) \le ln(\prod^n_{j=1} b_j) \tag{g}$$
From (f) we see that $f(x) = -ln(x)$ is a strictly decreasing function. Therefore, $-f(x) = ln(x)$ is a strictly increasing function. This implies along with (g) that,
$$ \prod^n_{i=1} a_i \le \prod^n_{j=1} b_j $$
Q.E.D
