Number array divided into several parts, genelize $a>b>c>d>0$ so $ab+cd>ac+bd$ to more numbers Now, we have an original number array: $$a_1 > a_2 > a_3 > ... > a_{mn} > 0$$, I wonder whether the following inequality is the truth, if so, could you give me the proof or some instruction?
inequality:
$$
a_1 \times a_2 \times \cdots \times a_n + a_{n+1} \times a_{n+2} \times \cdots \times a_{2n} + \cdots + a_{i \times n+1} \times a_{i\times n+2} \times \cdots \times a_{i\times n+n} + \cdots + a_{(m-1)  n+1} \times a_{(m-1)  n+2} \times \cdots \times a_{(m-1)  n+n} > a_1 \times a_{1+m} \times \cdots \times a_{1+(n-1)m} + a_2 \times a_{2+m} \times \cdots \times a_{2+(n-1)m} + \cdots + a_i \times a_{i+m} \times \cdots \times a_{i+(n-1)m} + \cdots + a_m \times a_{m+m} \times \cdots \times a_{m+(n-1)m}
$$
That is,
$$
\sum_{i=1}^m\prod_{j=1}^n a_{j+n(i-1)} > \sum_{i=1}^m\prod_{j=1}^n a_{i+m(j-1)}
$$
Namely, the left partitions the number array into $m$ parts each of which has $n$ numbers and the numbers in each part are next to each other in the original array. While, the right partitions the number array into $m$ parts each of which has $n$ numbers and the adjacent two numbers in each part have a distance of $m$ in original array.
For example, we have six numbers: $$a>b>c>d>e>f>0$$, we want to partition it into 3 blocks, each block has 2 numbers, namely $m=3, n=2$, I want to know whether $$ab+cd+ef>ad+be+cf $$and prove it. Certainly, this is easy. The question is I want to generalize it to $m\times n$ numbers divided into $m$ blocks.
Thank you!
 A: In general, given $a_1>a_2>a_3>...>a_{mn}>0$ and from all the ways to partition the number array into m parts each of which has n numbers, the maximum value of the sum of product of each part will be $\sum\limits_{i=1}^m\prod\limits_{j=1}^na_{n(i-1)+j}$.
To prove it, we need to find a series of inequality which shows that the sum of product of any other partition is smaller than this one.
First, given $a_k>a_l$ and sum=$a_k\prod\limits_{p}a_p+a_l\prod\limits_{q}a_q$. If we exchange $\prod\limits_{p}a_p$ and $\prod\limits_{q}a_q$, will the sum become larger? It depends on the products, if $\prod\limits_{p}a_p<\prod\limits_{q}a_q$, exchanging them will increase the sum.
Now we need to find a way to put $a_1,a_2...a_n$ together. First we find the part which contains $a_1$, simply call it part X. Then we find the part which contains $a_2$. If it's also part X, we look for $a_3$ and so on. If it's a different part, let's call it part Y.
We need to put $a_2$ into part X and increase the sum. Now we know that $a_1$ is the largest number, but we don't know about the others, so we cannot simply exchange $a_2$ and one number from X. To solve this problem, we have to exchange more numbers.
So we list all the 2n numbers from X and Y on a line,
　　　$a_1　...$　(numbers from X)
Large ---------- Small
　　　　$a_2　...$(numbers from Y)
And then we put a dot to the left of $a_1$. Then we move it to a position between $a_1$ and $a_2$, and then move it to the right of $a_2$, then move it rightward to skip another number and so on. Define $c(i)$ to be the number of numbers from X which are to the right of the dot after i moves, and $d(i)$ to be the number of numbers from Y which are to the left of the dot. At first we have $c(1)=(n-1)$ and $d(1)=0$, when i increase by 1, $c(i)$ decrease by 1 or $d(i)$ increase by 1. At last $c(2n)=0$ and $d(2n)=n$. We can see that $c(1)>d(1)$ and $c(2n)<d(2n)$, so that there must be a moment when $c(j)=d(j)$, say this number is $t$.

Now we can exchange these $t$ numbers. Define $X_l$ to be the set and $x_l$ to be the product of the numbers from X which are to the left of the dot, and $X_r$ to be the set and $x_r$ to be the product of the right ones, $Y_l$ to be the set and $y_l$ to be the product of the numbers from Y which are to the left of the dot, and $Y_r$ to be the set and $y_r$ to be the product of the right ones. 
$X_l$ and $Y_r$ each contains $n-t$ numbers, $X_r$ and $Y_l$ each contains $t$ numbers, and since each number from $X_l$ is larger than any number from $Y_r$ and each number from $Y_l$ is larger than any number from $X_r$, we have $x_l>y_r$ and $y_l>x_r$. It's easy to prove that $x_lx_r+y_ly_r<x_ly_l+x_ry_r$. Now we can exchange $Y_l$ and $X_r$ and the sum will increase.
Since $a_2$ is in $Y_l$, now we have moved it into part X. Then we do the same thing to the part which contains $a_3$ and so on, at last we will have $a_1, a_2, a_3...a_n$ in X. After this one is done, we start with the part which contains $a_{n+1}$ and do the same thing. At last the sum will be $\sum\limits_{i=1}^m\prod\limits_{j=1}^na_{n(i-1)+j}$ and that will be the maximum.
