Prove that $|A_1A_2|^2+|A_2A_3|^2+\ldots+|A_{n-1}A_n|^2+|A_nA_1|^2\leq 9R^2$. A polygon $A_{1}A_{2}...A_{n}$ has a circumscribed circle with radius  $R$. Prove $$|A_1A_2|^2+|A_2A_3|^2+\ldots+|A_{n-1}A_n|^2+|A_nA_1|^2\leq 9R^2.$$
 A: You can prove that by induction on $n$.
The base case $n = 3$ is the hardest one and has been settled in another
question.
Let's say you have proved the statement for all $k$-gons with $3 \le k < n$ where $n \ge 4$.
Consider any $n$-gon $A_1 A_2 \ldots A_n$ inscribed in a circle of radius $R$. Since the sum of exterior angles is $360^\circ$, at least one of the exterior angle $\;\le \frac{360}{n}^\circ \le 90^\circ$. Relabel vertices if necessary, we can assume the exterior angle at $A_2 \le 90^\circ$. The triangle $A_1A_2A_3$ will be a non-acute triangle and $|A_1A_2|^2 + |A_2A_3|^2 \le |A_1A_3|^2$. This leads to
$$|A_1A_2|^2 + |A_2A_3|^2 + |A_3A_4|^2 + \cdots + |A_nA_1|^2
\le  |A_1A_3|^2 + |A_3A_4|^2 + \cdots + |A_nA_1|^2$$
Apply induction assumption to the $(n-1)$-gon $A_1A_3A_4\ldots A_n$, one obtain RHS $\le 9R^2$ and we are done.
Update
It turns out the base case at $n = 3$ isn't that hard at all.
Choose a coordinate system where the circumcenter of triangle
$A_1A_2A_3$ is origin. i.e. $|\vec{A}_1| = |\vec{A}_2| = |\vec{A}_3| = R$.
We have
$$\begin{align} 
  &\; |A_1A_2|^2 + |A_2A_3|^2 + |A_3A_1|^2\\
= &\; |\vec{A}_1 - \vec{A}_2|^2 + |\vec{A}_2-\vec{A}_3|^2 + |\vec{A}_3-\vec{A}_1|^2\\
= &\; 2(|\vec{A}_1|^2 + |\vec{A}_2|^2 + |\vec{A}_3|^2) - 2(\vec{A}_1\cdot\vec{A_2} + \vec{A}_2\cdot\vec{A}_3 + \vec{A}_3\cdot\vec{A}_1)\\
= &\; 3(|\vec{A}_1|^2 + |\vec{A}_2|^2 + |\vec{A}_3|^2) -|\vec{A}_1+\vec{A}_2+\vec{A}_3|^2\\
\le &\; 3(|\vec{A}_1|^2 + |\vec{A}_2|^2 + |\vec{A}_3|^2)\\
= &\; 9R^2
\end{align}
$$
A: Assume $n\geq4$. The interior angle sum of your polygon $P$ is $(n-2)\pi$; hence at least one of these angles  satisfies $$\phi\geq{(n-2)\pi\over n}=\pi-{2\pi\over n}\geq{\pi\over2}\ .$$ Let  $\phi=\angle(A_{n-1}, A_n ,A_1)\geq{\pi\over2}$.  Then by the cosine theorem
$$|A_{n-1}A_1|^2=:c^2= a^2+b^2-2ab\cos\phi\geq|A_{n-1}A_n|^2+|A_nA_1|^2\ .$$
It follows that deleting the vertex $A_n$ from $P$ creates a polygon $P'$ with  cyclic sum $\sum_{cyc}|A_{k-1}A_k|^2$ at least as large. It therefore suffices to prove the claim for triangles. A simple proof is given in the update to achille hui's answer.
