Proof idea for sum of square / square of sum inequality This problem came up in a long proof and it is not clear to me how to show this. I tried to apply Cauchy-Schwarz but it is not tight enough. Any idea?
Let $\Delta$ be a matrix of real values with $B$ rows and $L$ columns.
Prove or disprove:
$(B+L-L\cdot B)\sum_{1\leq i\leq B\\1\leq j \leq L}\Delta^2_{i, j}
+ B\cdot\sum_{1\leq i\leq B\\1\leq j \leq L}\sum_{1\leq j'\leq L\\j'\neq j}\Delta^2_{i, j}\Delta^2_{i, j'}
+ L\cdot\sum_{1\leq i\leq B\\1\leq j \leq L}\sum_{1\leq i'\leq B\\i'\neq i}\Delta^2_{i, j}\Delta^2_{i', j}
\leq (\sum_{1\leq i\leq B\\1\leq j \leq L}\Delta_{i, j})^2
$

Example (B = 3, L = 3):
D = ( 1  2  0
     -4 -3 -5
      4  5  3)

-3 ⋅ 105 + 3 ⋅ 192 + 3 ⋅ (-84) = 9 ≤ 9

Example (B = 4, L = 3):
D = ( -9  -7  -6
       0   6  -3
       9  12  11
       1 -10  10)

-5 ⋅ 758 + 4 ⋅ 760 + 3 ⋅ (-612) = -2586 ≤ 196

 A: Finally, I found the solution. Evidently, inequalities such as Cauchy-Schwarz are not tight enough. Therefore, the inequality must be transformed so that it consists of positive terms only:
Subtracting the right-hand side form the left-hand side and multiplying by $-1$ yields:
$$
+ (B\cdot L + 1 - B - L)\sum_{\substack{1\leq i\leq B \\ 1\leq j\leq L}} \Delta^2_{i,j}\\
- (1 - B) \sum_{\substack{1\leq i\leq B \\ 1\leq j\leq L}} \sum_{\substack{1\leq j'\leq L \\ j'\neq j}} \Delta_{i,j} \cdot \Delta_{i,j'}\\
-(1 - L) \sum_{\substack{1\leq i\leq B \\ 1\leq j\leq L}} \sum_{\substack{1\leq i'\leq B \\ i'\neq i}} \Delta_{i,j} \cdot \Delta_{i',j}\\
+ \sum_{\substack{1\leq i\leq B \\ 1\leq j\leq L}}\sum_{\substack{1\leq i'\leq B \\ 1\leq j'\leq L \\ i'\neq i \\ j'\neq j}} \Delta_{i,j} \cdot \Delta_{i',j'} \geq 0
$$
This is equivalent to:
$$
\underset{i<i'}{\sum^B_{i=1}\sum^B_{i'=1}}\text{ }\underset{j<j'}{\sum^L_{j=1}\sum^L_{j'=1}} (\Delta_{i, j} - \Delta_{i, j'} - \Delta_{i', j} + \Delta_{i', j'})^2 \geq 0
$$
