Positive correlation with the sequence $\sqrt{ij}/2-\min(i,j)$ Is there a sequence of positive real numbers $x_1,\ldots,x_n$ for which
$$
\sum_{1\leq i,j\leq n}\left[\frac{\sqrt{ij}}{2}-\min(i,j)\right]x_ix_j> 0?
$$
 A: Alternative solution:
There exists such $n$ and a sequence of positive real numbers $x_1, x_2, \cdots, x_n$.
Let $A$ denote the matrix whose $(i,j)$-th entry is $\frac{\sqrt{ij}}{2}$.
Then, $A = uu^\mathsf{T}$ where
$u = \frac{1}{\sqrt{2}}[\sqrt{1}, \sqrt{2}, \cdots, \sqrt{n}]^\mathsf{T}$.
Let $B$ denote the matrix whose $(i,j)$-th entry is $\min(i, j)$.
Note that $B^{-1}$ is a symmetric tridiagonal matrix of the form
\begin{align}
B^{-1} =  \begin{pmatrix} 2 & -1 & 0 & \ldots & 0 & 0 \\ -1 & 2 & -1 & \ldots & 0 & 0 \\ 0 & -1 & 2 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \ldots & 2 & -1 \\ 0 & 0 & 0 & \ldots & -1 & 1 \end{pmatrix}.
\end{align}
(Note: The diagonal entries are $2, 2, \cdots, 2, 1$. The subdiagonal and superdiagonal entries are all $-1$. )
Let $v = B^{-1}u$.
Fact 1: $v_i > 0$ for $i=1, 2, \cdots, n$.
Fact 2: $u^\mathsf{T}B^{-1}u = \frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)}$.
Fact 3: $v^\mathsf{T}(A - B)v = (u^\mathsf{T}B^{-1}u)^2 - u^\mathsf{T}B^{-1}u$.
The proofs are easy and thus omitted.
Now, from Facts 1, 2, 3, if $u^\mathsf{T}B^{-1}u > 1$ or equivalently $\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} > 1$, then $v^\mathsf{T}(A - B)v > 0$.
It is easy to prove that there exists $n$ such that $\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} > 1$.
Indeed, by using
$\sqrt{k(k+1)} \le k + \frac{1}{2} - \frac{1}{16k}$ for $k \ge 1$, we have
$$\sum_{k=1}^{n-1} \sqrt{k(k+1)} \le \frac{n^2}{2} - \frac{1}{2} - \frac{1}{16}\sum_{k=1}^{n-1} \frac{1}{k}$$
which results in
$$\frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)} \ge \frac{1}{2} +  \frac{1}{16}\sum_{k=1}^{n-1} \frac{1}{k}.$$
The desired result follows.
For such $n$ and $v$, we have
$$\sum_{1\le i, j\le n} \left[\frac{\sqrt{ij}}{2} - \min(i, j)\right] v_iv_j = v^\mathsf{T}(A - B)v > 0.$$
Remark: Denote $F(n) = \frac{n^2}{2} - \sum_{k=1}^{n-1} \sqrt{k(k+1)}$. By Maple, $F(55) = 0.99974443 < 1$ and $F(56) = 1.00199623$.
