Double Summation - Cauchy-Schwarz for an Array

Question

From Steele's Cauchy-Schwarz Master Class I'm having a hard time understanding the solution problem 1.13:

Show that if {$a_{jk}$: 1 ≤ j ≤ m, 1 ≤ k ≤ n} is an array of real numbers then one has

$$m\sum_{j=1}^m\left(\sum_{k=1}^na_{jk}\right)^2 + n\sum_{k=1}^n\left(\sum_{j=1}^ma_{jk}\right)^2 \leq \left(\sum_{j=1}^m\sum_{k=1}^na_{jk}\right)^2 + mn\sum_{j=1}^m\sum_{k=1}^na_{jk}^2$$

The solution outline: "If we view {$x_{jk}$ : 1 ≤ j ≤ m, 1 ≤ k ≤ n} as a vector of length mn then Cauchy’s inequality and the one-trick splitting xjk = xjk · 1 imply the general bound

($\sum_{j=1}^m\sum_{k=1}^nx_{jk})^2$ ≤ mn $\sum_{j=1}^m\sum_{k=1}^nx_{jk}^2$

We apply this bound to $x_{jk}$ = $a_{jk}$ − $r_j$/n − $c_k$/m where $r_j$ = $\sum_{k=1}^na_{jk}$ and $c_k = \sum_{j=1}^ma_{jk}$ and if we set T = $\sum_{j=1}^m\sum_{k=1}^na_{jk}$

We are suppose to get $T^2$ for the left side of the bound I don't follow the summation for this part.

Answer 1

I think it means that we need to prove $$m\sum_{j=1}^m\left(\sum_{k=1}^na_{jk}\right)^2 + n\sum_{k=1}^n\left(\sum_{j=1}^ma_{jk}\right)^2 \leq \left(\sum_{k=1}^n\sum_{j=1}^ma_{jk}\right)^2 + mn\sum_{k=1}^n\sum_{j=1}^ma_{jk}^2,$$ which is just $$\sum_{1\leq j<s\leq m,1\leq k<l\leq n}(a_{jk}-a_{jl}-a_{sk}+a_{sl})^2\geq0.$$ For example, for $m=2$ and $n=3$ we need to prove that $$2((a+b+c)^2+(x+y+z)^2)+3((a+x)^2+(b+y)^2+(c+z)^2)\leq$$ $$\leq(a+b+c+x+y+z)^2+6(a^2+b^2+c^2+x^2+y^2+z^2)$$ or $$(a-b-x+y)^2+(a-c-x+z)^2+(b-c-y+z)^2\geq0.$$