# Double Summation - Cauchy-Schwarz for an Array

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.

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 $$\sum_{1\leq j For example, for $$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.$$