Proving the following inequality (positive def. matrix) I'm trying to prove (or disprove) the following:
$$ \sum_{i=1}^{N} \sum_{j=1}^{N} c_i c_j K_{ij}  \geq  0$$
where $c \in \mathbb{R}^N$, and $K_{ij}$ is referring to a kernel matrix:
$$K_{ij} = K(x_i,x_j) = \frac{\sum_{k=1}^{N} \min(x_{ik}, x_{jk})}{\sum_{k=1}^{N} \max(x_{ik}, x_{jk})}$$
Here, $x \in \mathbb{R}^N \geq 0$.
I'm basically trying to prove that $K_{ij}$ is a positive definite matrix, so I can use it as a Kernel, but I'm really stuck trying to work with $\max$
Edit: the function I'm refering to is:
$$K(u,v) = \frac{\sum_{k=1}^{N} \min(u_{k}, v_{k})}{\sum_{k=1}^{N} \max(u_{k}, v_{k})}$$
where $u, v \in \mathbb{R}^N \geq 0$

Fix $x_i\in\mathbb{R}^n$, $i = 1, 2, \ldots, N$. We will assume without loss of generality that no $x_i$ is identically $0$. Define $N\times N$ matrices $A = (\sum_{k=1}^n\min(x_{i(k)}, x_{j(k)}))$  and $B = (\sum_{k=1}^n\max(x_{i(k)}, x_{j(k)}))$, where $x_{(k)}$ denotes the $k$th coordinate of $x$. Note that $K = A\odot B^{\odot-1}$ where $\odot$ denotes the Hadamard product and $B^{\odot-1}$ is the Hadamard inverse (entrywise reciprocal) of $B$. By the Schur product theorem, it suffices to show that $A$ and $B^{\odot-1}$ are positive definite. We will use the fact that a positive linear combination of positive definite matrices is positive definite.
To show that $A$ is positive definite, note that $A$ can be written as the sum $A = \sum_{k=1}^n A_k$ with $A_k = \min(x_{i(k)}, x_{j(k)})$. It suffices to show that e.g. $A_1$ is positive definite. For $i \in [N]$, let $y_i = x_{i(1)}$. By conjugating $A_1$ by a permutation matrix, we may assume without loss that $y_1\leq y_2\ldots \leq y_N$. For $i \in [N]$, let $f_i\in\mathbb{R}^N$ denote the vector with $f_{i(j)} = 0$ for $j < i$ and $f_{i(j)} = 1$ for $j \geq i$. Then, setting $y_0 = 0$,
\begin{equation}A_1 = \sum_{i=1}^N(y_i - y_{i-1})f_if_i^t \geq 0.
\end{equation}
We now show that $B^{\odot-1}$ is positive definite. By scaling, we may assume that $x_{i(j)} \in [0, 1/n]$ for all $i$ and $j$. Using the identity $1/x = \sum_{i=0}^{\infty}(1-x)^i$ valid for $x\in (0, 2)$, we may write $B^{\odot-1} = J + \sum_{i=1}^{\infty} (J-B)^{\odot i}$, where $J=f_1f_1^t$ denotes the all ones matrix. Now
\begin{equation}(J-B)_{ij} = 1 - \sum_{k=1}^n\max(x_{i(k)}, x_{j(k)}) = \sum_{k=1}^n \min(\frac{1}{n}-x_{i(k)}, \frac{1}{n}-x_{j(k)}).
\end{equation} The above argument that showed that $A$ is positive definite now shows that $J-B$ is positive definite (by replacing $x_i$ with $x_i' = \frac{1}{n}f_1 - x_i$). Finally, the Schur product theorem and the fact that positive definite matrices are closed under positive linear combinations show that $B^{\odot-1}$ is positive definite.
