Impossible Decagon

Let $a_i$ be the $i$ th element of $A$ where $1\leq i \leq 10$ .

$S(A)=\displaystyle \sum_{i=1}^{10} a_i=a_1+a_2+...+a_9+a_{10}$

WLOG, we let $a_1 \leq a_2 \leq ... \leq a_{10}$

For something to be a decagon, no side length can be greater than or equal to the sum of the other nine sides. Our set $A$ must fail this condition, therefore we have:

$a_1+a_2+...+a_9 \leq a_{10}$

$a_1+a_2+...+a_9+a_{10} \leq 2a_{10}$

$S(A) \leq 2a_{10}$

By the Least Upper Bound Property, any non-empty set of real numbers that has an upper bound necessarily has a least upper bound. If $2a_{10}$ is the upper bound for $S(A)$ , to find the least upper bound we must minimize $a_{10}$ .

$1 \leq a_i$ $\forall i \Longrightarrow 9\leq \displaystyle \sum_{i=1}^9 a_i \leq a_{10}$

$9+a_{10} \leq S(A) \leq 2a_{10}$

We now have $a_{10}=9$ as a minimum, meaning that the least upper bound for $S(A)$ is $2(9)=18$ . However, our last inequality gives us a lower bound of $9+a_{10}=9+9=18$ . Due to the Squeeze Theorem, if $X \leq Y \leq Z$ and $X=Z$ , then $X=Y=Z$ . Since $18\leq S(A) \leq 18$ , we have our minimum value $S(A) = \boxed{18}$ .

The proof of the generalization is quite similar, but I'll leave the fun of that to you ;)