If $a_1,a_2,a_3,...,a_n$ are the side lengths of $A_1A_2A_3...A_n$ convex polygon,then$\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}$ is If $a_1,a_2,a_3,...,a_n$ are the side lengths of $A_1A_2A_3...A_n$ convex polygon,then$\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}$ is 
$(A)>\frac{1}{n-1}\hspace{1cm}(B)<\frac{1}{n-1}\hspace{1cm}(C)>\frac{1}{n}\hspace{1cm}(D)<\frac{1}{n}$
$\frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}=\frac{a^2_1}{a^2_n}+\frac{a^2_2}{a^2_n}+\frac{a^2_3}{a^2_n}+......+\frac{a^2_{n-1}}{a^2_n}>(n-1)\frac{a^2_1}{a^2_n}\frac{a^2_2}{a^2_n}\frac{a^2_3}{a^2_n}......\frac{a^2_{n-1}}{a^2_n}$

and then no idea how to proceed.Any hints,please?
 A: Take a regular polygon with $n$ sides, which is convex. Add an edge an one corner, of lenght sufficiently small. Then
$$
\sum_{i=1}^{n-1}\left(\frac{a_i}{a_n}\right)^2
$$
can be arbitrarily large. On the other hand, it follows by induction on triangle inequality that $a_n<\sum_{i=1}^{n-1}a_i$. Without loss of generality we set $\sum_{i=1}^{n-1}a_i=1$. This implies that
$$
\sum_{i=1}^{n-1}\left(\frac{a_i}{a_n}\right)^2>\sum_{i=1}^{n-1}a_i^2\ge\frac{1}{n-1}
$$
where the last inequality follows by AM-GM.
A: Hint Let $a_1,a_2,a_3,...,a_n=x$ be are the side lengths of $A_1A_2A_3...A_n$ a convex polygon, the idea is to modify $x$. for instance we can have the inequality $0\leq x\leq a_1+\cdots+a_{n-1}=A\tag 1$ we can see that $x$ can take values very close to $A$ (see the situation explained in case: $1$) and it can take also very small values (see the situation explained in case: $2$). So we can say that $(1)$ is sharp. Here are two graphs to explain that:


Now we can deduce from this that :
$$\frac{1}{n-1} \leq \frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{(a_1+a_2+\cdots+a_{n-1})^2}\leq \frac{a^2_1+a^2_2+a^2_3+....+a^2_{n-1}}{a^2_n}\leq +\infty $$
using the inequality (coming from Cauchy–Schwarz inequality) $$(a_1+a_2+\cdots+a_{n-1})^2\leq (n-1)(a^2_1+a^2_2+a^2_3+....+a^2_{n-1}) $$
