Maximize it (2)

$(a+b+c+d)^2 = a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd) \\ \implies ab+bc+cd+da = \dfrac{(a+b+c+d)^2-a^2-b^2-c^2-d^2}{2}-(ac+bd)$ .

We can maximize $ab+bc+cd+da$ by minimizing $ac+bd$ . Checking the three distinct values of $ac+bd$ , which are

$1 \times 2 + 3 \times 4$ ,

$1 \times 3 + 2 \times 4$

and $1 \times 4 + 2 \times 3$ ,

we see that the smallest one is $1 \times 4 + 2 \times 3 = 10$ .

Thus, the largest possible value of $ab+bc+cd+da$ is $\dfrac{(1+2+3+4)^2-1^2-2^2-3^2-4^2}{2}-10 = \space \boxed{25}$

$ab+bc+cd+da=(a+c)(b+d)$

We know that the sum $a+b+c+d=10$ . By AM-GM Inequality, $\dfrac{(a+c)+(b+d)}{2} \geq \sqrt{(a+c)(b+d)}$ . Upon evaluating the inequality, we find that the maximum value of $(a+c)(b+d)$ is $5$ .

$ab + ac + ad + bc + bd + cd$ = 35 Our purpose is to minimize $ac + bd$ and it happen when $a=1,b=2,c=4,d=3$

$35-10=25$