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.

Solution 1:

Here is a derivation using the calculus of variations.

Given that $$ \sum_{k=1}^na_k^2=1\tag{1} $$ we want to find the maximum of $$ \sum_{k=1}^na_ka_{k-1}\tag{2} $$ where $a_0=a_{n+1}=0$.

We want to find $a_k$ so that $(2)$ is stationary, that is, $$ \begin{align} 0 &=\sum_{k=1}^na_k\,\delta a_{k-1}+a_{k-1}\,\delta a_k\\ &=\sum_{k=1}^n(a_{k-1}+a_{k+1})\,\delta a_k\tag{3} \end{align} $$ for all $\delta a_k$ so that $(1)$ is constant: $$ 0=\sum_{k=1}^na_k\,\delta a_k\tag{4} $$ For both the $a_{k-1}+a_{k+1}$ in $(3)$ and the $a_k$ in $(4)$ to be perpendicular to all the same $\delta a_k$, we must have some $\lambda$ so that $$ a_{k+1}+a_{k-1}=\lambda a_k\tag{5} $$ Solving $(5)$, so that $a_0=a_{n+1}=0$, using the standard methods for linear recurrences yields $$ a_k=r\sin\left(\frac{\pi mk}{n+1}\right)\tag{6} $$ for $m\in\mathbb{Z}$.

We can compute $r$ using $(1)$ and $(6)$: $$ \begin{align} \sum_{k=1}^na_k^2 &=\sum_{k=1}^nr^2\sin^2\left(\frac{\pi mk}{n+1}\right)\\ &=-\frac{r^2}{4}\sum_{k=1}^n\left[\exp\left(\frac{2\pi imk}{n+1}\right)+\exp\left(\frac{-2\pi imk}{n+1}\right)-2\right]\\ &=-\frac{r^2}{4}(-2n-2)\ \Big[m\not\equiv0\bmod{(n+1)}\Big]\\ &=r^2\frac{n+1}{2}\ \Big[m\not\equiv0\bmod{(n+1)}\Big]\tag{7} \end{align} $$ We cannot satisfy $(1)$ when $m\equiv0\bmod{(n+1)}$; therefore, we require $m\not\equiv0\bmod{(n+1)}$, in which case, $$ r=\sqrt{\dfrac2{n+1}}\tag{8} $$

Finally, we need to compute $(2)$ for each $m\not\equiv0\bmod{(n+1)}$: $$ \begin{align} &\frac2{n+1}\sum_{k=1}^n\sin\left(\frac{\pi mk\vphantom{()}}{n+1}\right)\sin\left(\frac{\pi m(k-1)}{n+1}\right)\\ &=\frac1{n+1}\sum_{k=1}^n\left[\cos\left(\frac{\pi\vphantom{()}m}{n+1}\right)-\cos\left(\frac{\pi m(2k-1)}{n+1}\right)\right]\\ &=\frac{n}{n+1}\cos\left(\frac{\pi m}{n+1}\right)+\frac1{n+1}\cos\left(\frac{\pi m}{n+1}\right)\\ &=\cos\left(\frac{\pi m}{n+1}\right)\tag{9} \end{align} $$ The largest that $(9)$ can be if $m\not\equiv0\bmod{(n+1)}$ is $\cos\left(\frac\pi{n+1}\right)$.

Therefore, by homogeneity, $$ \frac{\displaystyle\sum_{k=1}^na_ka_{k-1}}{\displaystyle\sum_{k=1}^na_k^2}\le\cos\left(\frac\pi{n+1}\right)\tag{10} $$

Solution 2:

I don't get the result you are looking for, a mistake may have crept in somewhere, I'll get back tot he question tomorrow morning. What I try to do is to find the spectrum of the matrix $D_n$ defined below. The solution uses the Tchebychev polynomials.

Edit. I made a mistake when identifying (the rescaled version of) the $\chi_n$ with the Tchebychew polynomials: the second polynomial $\tau_2=\frac12\chi_2(-2X)$ is actually equal to $2X^2-\frac12$ and not $T_2=2X^2-1$. I'm not sure I can rescue the proof.

Consider the real symmetric matrix $$D_n=\begin{pmatrix} 0&1\\ 1&0&1\\ &1&0\\ &&&\ddots\\ &&&&0&1\\ &&&&1&0 \end{pmatrix}$$ Then the equation above is equivalent to $$\langle D_nX\,|\,X\rangle\leq2|X|^2\cos\left(\frac{\pi}{n+1}\right)$$ for all nonzero vectors $X\in\mathbb{R}^{n}$, where $|X|^2=\langle X\,|\,X\rangle$. Because $D_n$ is diagonalizable in an orthonormal basis, it is enough to show that all the eigenvalues of $D_n$ are of absolute value less or equal than $2\cos\left(\frac{\pi}{n+1}\right)$.

The characteristic polynomial $\chi_n(X)=\det(D_n-XI)$ of $D_n$ the relations $\chi_1=-X$, $\chi_2=X^2-1$ (and $\chi_3=-X^3+2X$) and an easy manipulation of determinants shows that for all $n\geq 2$, $$\chi_n+X\chi_{n-1}+\chi_{n-2}=0$$

Compare this to the linear recursion satisfied by the Tchebychew polynomials of the first (and second) kind : $$T_n-2XT_{n-1}+T_{n-2}=0$$ The sequence $(\tau_n)=(\frac12\chi_n(-2 X))_n$ satisfies the recursion formula above, and the first two terms agree with the first two Tchebychew polynomials, for $\tau_1=X=T_1$ and $\tau_2=2X^2-1=T_2$ so that for all $n\geq 0$, $\tau_n=T_n$ and $$\chi_n(X)=2T_n\left(-\frac12 X\right)$$ Since the roots of $T_n$ are the $\cos\left(\frac{2k+1}{2n}\pi\right)$ for $k=0,\dots,n-1$, we get the roots of $\chi_n$, i.e. $$\mathrm{Spec}(D_n)=\left\lbrace 2\cos\left(\frac{2k+1}{2n}\pi\right)\left|\right. k=0,\dots,n-1\right\rbrace$$ And so all eigenvalues are (in abolute value) at most $2\cos\left(\frac{\pi}{2n}\right)$.