A game of a,b,c and d

I have 2 methods to solve this question. One is algebraic and the other is geometric.

$\Rightarrow \!\,$ Method 1 - Algebraic method:

For all real numbers $x$ and $y$ , we have $(x-y)^2 ≥0$ .

So $x^{2} +y^{2} ≥ 2xy$ , hence $(x + y)^2 ≥4xy$ and $xy ≤ (x + y)^2 /4$ .

Letting $x = a+c$ and $y = b+d$ gives $(a+c)(b+d) ≤ (a+b+c+d)^2 /4$ . So $ab + bc + cd + da≤63^{2}/4 = 3969/4 = 992.25$ .

Since $a,b,c,d$ are positive integers, the last inequality can be written as $ab+ bc + cd + da≤ 992$ . Hence $ab + bc + cd ≤ 992-da≤ 991$ . It remains to show that 991 is achievable.

Suppose $ab+bc+cd = 991$ and $a = d = 1$ . Then $(1 + b)(1 + c) = 992 = 25 ×31$ . So $b = 30$ and $c = 31$ is a solution.

Thus the maximum value of

$\large ab+bc+cd = 991$ .

$\Rightarrow \!\,$ Method 2 - Geometric method