Given any positive integer $n$, let $t(n)$ be the smallest real number such that for any real numbers $x_1,\dots,x_n$, the following inequality holds $$ \sum\limits_{k=1}^n (x_1+\dots+x_k)^2 \leqslant t(n) \sum\limits_{i=1}^n x_i^2.$$ Please give an expression for $t(n)$.
Attempt:
The problem is equivalent to calculating the maximum of $\sum\limits_{k=1}^n (x_1+\dots+x_k)^2$ on the unit $n$-sphere $\sum\limits_{i=1}^n x_i^2=1$. Let
$$ A := \begin{pmatrix}
n & n-1 & n-2 & \cdots & 1\\
n-1 & n-1 & n-2 & \cdots & 1\\
n-2 & n-2 & n-2 & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1
\end{pmatrix}$$
Then
$$\sum\limits_{k=1}^n (x_1+\cdots+x_k)^2=xAx^T$$
where $x=(x_1,\dots,x_n)$. According to Lagrange multipliers, we know the problem is equivalent to calculating the maximum eigenvalue of $A$. Then I'm stuck. The book says that the answer is
$$\frac{1}{4\sin^2\left(\frac{\pi}{4n+2}\right)}.$$
It is really amazing. I would appreciate it if someone could please give me some hints.
Best Answer
Hint. Note that the inverse of $A$ is $$B_n=\begin{pmatrix} 1 & -1 \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ &&&\ddots\\ &&& -1 & 2 &-1\\ &&&& -1 & 2 \end{pmatrix}.$$ Then the characteristic polynomial $P_n$ of the tridiagonal matrix $B_n$ satisfies the recurrence relation: $$P_{n}(x)=P_{n-1}(x)(x-2)-P_{n-2}(x)$$ with $P_1(x)=x-1$ and $P_0(x)=1$.
See also the related question Eigenvalues of tridiagonal symmetric matrix with diagonal entries $2$ and subdiagonal entries $1$