Largest Sum

Let $\displaystyle \sum_{i=1}^3 x_i = X$ and $\displaystyle \sum_{j=1}^3 y_j =Y$

$X + Y = 2 + 3 + 5 + 7 + 11 + 13 = 41$

$\displaystyle \sum_{i=1}^3 \displaystyle \sum_{j=1}^3 x_i y_j = \displaystyle \sum_{i=1}^3 x_i \times \displaystyle \sum_{j=1}^3 y_j = X \times Y$

But $4XY = (X + Y)^2 - (X-Y)^2 \implies XY$ is maximized when $(X-Y)^2$ is minimum. $\implies (X-Y)^2 = 1$ ... (0 is Not possible as X, Y are integers adding up to 41)

This is achievable at $X = 20, Y =21$ or $X = 21, Y = 20$ (e.g. $x_i = 2, 7, 11, y_j = 3, 5, 13$ )

So ans. $20 \times 21 = \boxed{420}$

Let $x=x_1+x_2+x_3, y=y_1+y_2+y_3$ , then $x+y=2+3+5+7+11+13=41$ , and the sum in the problem is: $xy = \frac{1}{4}[(x+y)^2-(x-y)^2] = \frac{1}{4}[41^2 - (x-y)^2]$

To maximize $xy$ , $x$ and $y$ need to be as close as possible, so we can let $x=2+7+11=20$ and $y=3+5+13=21$ . Hence the largest value of the sum is $xy=420$ .

The sum we are looking for is $\sum_{i=1}^3 (x_i y_1+x_i y_2+x_i y_3)=(x_1+x_2+x_3)(y_1+y_2+y_3)$ .

Now, notice that the sum $x_1+x_2+x_3+y_1+y_2+y_3=2+3+5+7+11+13=41$ . Let $x_1+x_2+x_3=a$ and $y_1+y_2+y_3=b$ .

We know that $a+b=41$ , and want to maximize $ab$ . The maximum occurs at $a=20, b=21$ , or vice versa*

To show that this is attainable, let $(x_1, x_2, x_3, y_1, y_2, y_3)=(2, 7, 11, 3, 5, 13)$ . It follows that $a=2+7+11=20$ and $b=3+5+13=21$ , so $ab=\boxed{420}$

*The maximum actually occurs at $(a, b)=(\frac{41}{2} , \frac{41}{2})$ , but $a, b$ must be integers. (Do you see why?)