An m × n array A of real numbers is a Monge array if for all i, j , k and l such that 1 ≤ i < k ≤ m and 1 ≤ j < l ≤ n, we have A[i,j] + A[k,l] ≤ A[i,l] + A[k,j] In other words, whenever we pick two rows and two columns of a Monge array and consider the four elements at the intersections of the rows and the columns, the sum of the upper-left and lower-right elements is less than or equal to the sum of the lower-left and upper-right elements. For example, the following array is Monge:
10 17 24 11 45
36 75 17 22 28
13 44 33 66 13
16 22 6 32 19
51 28 29 34 17
37 21 53 23 23
24 7 23 6 34
Prove that an array is Monge if and only if for all i = 1, 2, 3...m−1 and j = 1, 2,...n−1 we have A[i,j] + A[i + 1,j + 1] ≤ A[i,j + 1] + A[j + 1,i]
HINT: The crucial idea for the induction step is that if we have an array
$$\begin{bmatrix}a&b&c\\d&e&f\end{bmatrix}$$
such that $a+e\le b+d$ and $b+f\le c+e$, then $a+f\le c+d$. Note that the first two inequalities can be rewritten as $a-d\le b-e$ and $b-e\le c-f$.
(By the way, your array isn’t Monge: $10+75=85>53=17+36$ in the upper lefthand corner.)