12th Czech-Polish-Slovak Mathematics Competition

Let $V=ab+bc+cd+da$. We will find the maximum value of $$V^2=(a+c{)}^2(b+d{)}^2=(a^2+c^2+2ac)(b^2+d^2+2bd).(1)$$ 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 $$\begin{array}{rl}{V}^2& =\left(10-b^2-d^2+\frac{8}{bd}\right)(b^2+d^2+2bd)=\\ & =10(b^2+d^2)+20bd+\frac{8(b^2+d^2)}{bd}+16-(b^2+d^2{)}^2-2bd(b^2+d^2).\end{array}(2)$$

Let $P=b^2+d^2$ and $Q=bd$; then $P\ge 2Q$ and $Q\ge 2$. Thus (2) becomes $$\begin{array}{rl}{V}^2& =10P+20Q+\frac{8P}{Q}+16-P^2-2PQ\\ & =-(P^2-10P+25)+\left(41-2PQ+20Q+\frac{8P}{Q}\right)\\ & =-(P-5{)}^2+\left[P\left(\frac{8}{Q}-2Q\right)+41+20Q\right].\end{array}$$ 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 $\sqrt{82}$.