Presumably the indices $i$ in $A_i$ are taken mod $n$,
so "$A_{n+1}$" is to be identified with $A_1$.
This must be a known isoperimetric inequality,
but it's easier to prove than to find in the literature.

Fix the area $\cal A$ of the $n$-gon. By a standard compactness argument
there exists an $n$-gon $A_1 A_2 \ldots A_n$ that minimizes
$\sum_{i=1}^n (A_i A_{i+1})^2$.
We first show that this polygon is convex (but possibly with
$A_{i-1} A_i A_{i+1}$ collinear for some $i$).
Indeed if it is not we can replace it by the convex hull,
with each side $A_j A_k$ of the convex hull
divided into $k-j$ equal subsegments; this both increases the area
and decreases $\sum_{i=1}^n (A_i A_{i+1})^2$,
so we can shrink the polygon back to area $\cal A$
and make the sum of its sides' squares even smaller.

Given convexity, fix all but one of the vertices, say $A_2$.
Then $A_2$ is limited to a line parallel to $A_1 A_3$,
and we readily see (as by choosing coordinates that make $A_1,A_3 = (\pm 1, 0)$ )
that $(A_1 A_2)^2 + (A_2 A_3)^2$ is minimized when
$(A_1 A_2) = (A_2 A_3)$. Thus the minimizing $n$-gon
has all sides equal, say with each $(A_i A_{i+1}) = s$;
and then $ns^2$ is minimized when $ns$ is $-$ but $ns$ is the perimeter,
and the usual isoperimetric inequality for $n$-gons
then finishes the proof that the area-$\cal A$ polygon
with the smallest $\sum_{i=1}^n (A_i A_{i+1})^2$ is regular.