Czech-Polish-Slovak Mathematics Competition

Let $V = ab + bc + cd + da$. We will find the maximum value of
$$V^2 = (a+c)^2(b+d)^2 = (a^2+c^2+2ac)(b^2+d^2+2bd). \quad (1)$$
All the given expressions do not change under the simultaneous replacement of $a$ by $b$, $b$ by $c$, $c$ by $d$ and $d$ by $a$. Since $ac \cdot bd = 4$, at least one of the numbers $ac$ and $bd$ is at least $2$. We may assume $bd \ge 2$.
Simple manipulations yield $ac = 4/bd$ and $a^2 + c^2 = 10 - b^2 - d^2$. We plug these expressions into (1) and get
$$\begin{aligned} V^2 &= \left(10 - b^2 - d^2 + \frac{8}{bd}\right) (b^2 + d^2 + 2bd) = \\ &= 10(b^2 + d^2) + 20bd + \frac{8(b^2 + d^2)}{bd} + 16 - (b^2 + d^2)^2 - 2bd(b^2 + d^2). \end{aligned} \quad (2)$$

Let $P = b^2 + d^2$ and $Q = bd$; then $P \ge 2Q$ and $Q \ge 2$. Thus (2) becomes
$$\begin{align*} V^2 &= 10P + 20Q + \frac{8P}{Q} + 16 - P^2 - 2PQ \\ &= -(P^2 - 10P + 25) + \left(41 - 2PQ + 20Q + \frac{8P}{Q}\right) \\ &= -(P - 5)^2 + \left[P\left(\frac{8}{Q} - 2Q\right) + 41 + 20Q\right]. \end{align*}$$
Obviously, $-(P-5)^2 \le 0$. The condition $Q \ge 2$ implies $8/Q - 2Q \le 8/2 - 2 \cdot 2 = 0$, which means the expression in the brackets is linear in $P$ with non-positive slope and it achieves its maximum for the smallest possible $P$. By $P \ge 2Q$ we obtain
$$V^2 \le 2Q \left( \frac{8}{Q} - 2Q \right) + 41 + 20Q = -4Q^2 + 20Q + 57 = -(2Q - 5)^2 + 82 \le 82.$$

Finally, we shall show that there are positive real numbers $a$, $b$, $c$, $d$ such that $V = \sqrt{82}$. The equality occurs if $P = 2Q = 5$, which is true for $b = d = \frac{1}{2}\sqrt{10}$. The numbers $a$ and $c$ satisfy $a^2 + c^2 = 5$, $ac = 8/5$. Thus
$$\{a, c\} = \left\{ \frac{\sqrt{41}-3}{2\sqrt{5}}, \frac{\sqrt{41}+3}{2\sqrt{5}} \right\} .$$

Answer. The maximum possible value of $ab + bc + cd + da$ is $\sqrt{82}$.