<s>[INST] How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$ 
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
  $$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$

I think this result is interesting.
When $n=2$, clearly
$$\dfrac{a_{1}a_{2}}{a^2_{1}+a^2_{2}}\le\dfrac{1}{2}=\cos{\dfrac{\pi}{3}}=\dfrac{1}{2}$$
When $n=3$,
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}}{a^2_{1}+a^2_{2}+a^2_{3}}\le\dfrac{\sqrt{2}}{2}.$$
This is true, because
$$a^2_{1}+\dfrac{1}{2}a^2_{2}\ge \sqrt{2}a_{1}a_{2},$$
$$\dfrac{1}{2}a^2_{2}+a^2_{3}\ge\sqrt{2}a_{2}a_{3}.$$
But for general $n$ I cannot prove it. 
 [/INST] For $t \in (0,\frac{\pi}{n})$, we have $\sin kt > 0$, for $k \in 1,\cdots,n-1$.
Thus from AM-GM inequality, $\dfrac{\sin (k+1)t}{\sin kt}a_k^2+\dfrac{\sin kt}{\sin (k+1)t}a_{k+1}^2 \ge 2a_ka_{k+1}$,
with equality iff $a_{k+1}\sin kt = a_k\sin (k+1)t$.
Summing over we have, $\displaystyle \sum\limits_{k=1}^{n-1} \left(\dfrac{\sin (k+1)t}{\sin kt}a_k^2+\dfrac{\sin kt}{\sin (k+1)t}a_{k+1}^2\right) \ge 2\sum\limits_{k=1}^{n-1}a_ka_{k+1}$
$\displaystyle \implies 2a_1^2\cos t + \sum\limits_{k=2}^{n} \left(\dfrac{\sin (k-1)t}{\sin kt}+\dfrac{\sin (k+1)t}{\sin kt}\right)a_k^2 \ge \dfrac{\sin (n+1)t}{\sin nt}a_n^2 + 2\sum\limits_{k=1}^{n-1}a_ka_{k+1}$
$\displaystyle \implies 2\cos t\sum\limits_{k=1}^{n}a_k^2 \ge \dfrac{\sin (n+1)t}{\sin nt}a_n^2 + 2\sum\limits_{k=1}^{n-1}a_ka_{k+1}$
Putting, $t = \dfrac{\pi}{n+1}$, $\sin (n+1)t = 0$.
Therefore, $\displaystyle \cos \dfrac{\pi}{n+1}\sum\limits_{k=1}^{n}a_k^2 \ge \sum\limits_{k=1}^{n-1}a_ka_{k+1}$ as desired.
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