62nd Belarusian Mathematical Olympiad

The sum of the diagonals $A_1{A}_3$, $A_3{A}_5$, $A_5{A}_7$, $A_7{A}_2$, $A_2{A}_4$, $A_4{A}_6$, $A_6{A}_1$ is obviously greater than the perimeter $P$ of the heptagon (see Fig. 1).

Fig. 1

Similarly, the sum of diagonals $A_1{A}_4$, $A_4{A}_7$, $A_7{A}_3$, $A_3{A}_6$, $A_6{A}_2$, $A_2{A}_5$, $A_5{A}_1$ is greater than $P$.

Hence the sum of all diagonals is greater than $2P$.

On the other hand, Fig. 2 shows that the constant $c=2$ cannot be improved.

Fig. 2