3 is correct since $$\sum^{n}_{i=1} (a_i + a_{i+2} -2a_{i+1} )^2 \ge 0,$$so an example of arithmetic sequence may show 3 is best.

BlueCloud12 wrote: so an example of arithmetic sequence may show 3 is best. Unfortunately, this is not how it works since any periodic arithmetic sequence trivially is constant and a constant sequence is not a counterexample at all... Indeed, for fixed $n$, the constant $C=3$ is not at all optimal. For instance, when $n=4$ the inequality is true for all $C \ge 1$, easily. So we need an "asymptotic counterexample" which can be given

Remark: For fixed $n$, the optimal constant $C_n$ is $1+2\cos(2\pi/n)$.

The inequality is equivalent to \[ \frac{C-1}{2} \sum_{i=1}^{n} (a_i-a_{i+1})^2 \ge \sum_{i=1}^{n} (a_i-a_{i+1})(a_{i+1}-a_{i+2}). \]Let $b_i=a_i-a_{i+1}$, then $\sum_{i=1}^{n} b_i=0$. Put $\frac{C-1}{2}=\cos \frac{2\pi}{n}$, it becomes \[ \cos \frac{2\pi}{n} \sum_{i=1}^{n} b_i^2 \ge \sum_{i=1}^{n} b_ib_{i+1}. \]This is well-known as Ky-Fan's inequality, which is also equivalent to the isoperimetric inequality.

NTstrucker wrote: The inequality is equivalent to \[ \frac{C-1}{2} \sum_{i=1}^{n} (a_i-a_{i+1})^2 \ge \sum_{i=1}^{n} (a_i-a_{i+1})(a_{i+1}-a_{i+2}). \]Let $b_i=a_i-a_{i+1}$, then $\sum_{i=1}^{n} b_i=0$. Put $\frac{C-1}{2}=\cos \frac{2\pi}{n}$, it becomes \[ \cos \frac{2\pi}{n} \sum_{i=1}^{n} b_i^2 \ge \sum_{i=1}^{n} b_ib_{i+1}. \]This is well-known as Ky-Fan's inequality, which is also equivalent to the isoperimetric inequality. Ah I did think the last inequality would be well-known, but I did not know the name. Could you maybe give a reference where it is referred as Ky-Fan's inequality? Also I'm wondering why it is equivalent to the isoperimetric inequality.

Case 1 very easy

hakN wrote: Let $n$ be some positive integer and $a_1 , a_2 , \dots , a_n$ be real numbers. Denote $$S_0 = \sum_{i=1}^{n} a_i^2 , \hspace{1cm} S_1 = \sum_{i=1}^{n} a_ia_{i+1} , \hspace{1cm} S_2 = \sum_{i=1}^{n} a_ia_{i+2},$$where $a_{n+1} = a_1$ and $a_{n+2} = a_2.$ 2. Show that $3$ is the minimum value of $C$ such that for any $n$ and $a_1 , a_2 , \dots , a_n,$ there holds $C(S_0 - S_1) \geq S_1 - S_2$. DFT kills these problems. \begin{align*}CS_0+S_2-(C+1)S_1=\sum_p\left(C+\cos\frac{4p\pi}n-(C+1)\cos\frac{2p\pi}n\right)|f(p)|^2.\end{align*}By $C=2\cos\frac{2\pi}n+1\le 2\cos\frac{2p\pi}n+1=\frac{\cos\frac{2p\pi}n-\cos\frac{4p\pi}n}{1-\cos\frac{2p\pi}n},$ $$C+\cos\frac{4p\pi}n-(C+1)\cos\frac{2p\pi}n\ge 0.$$done.$\Box$

EthanWYX2009 wrote: DFT kills these problems. Pardon me, could anyone explain what DFT is?

Discrete Fourier Transform