Inequality on numerical range of inverse of kernel matrix

Let $$k(.,.)$$ be a function that takes two vectors as input and outputs a scalar as follows \begin{align} \mathcal{k}(x,y) = \exp\left(-\frac{\|x-y\|_2^2}{2}\right) \end{align} where $$\|x\|_2$$ denotes the $$2-$$norm. Now, let $$x_1,\dots,x_m$$ be $$m$$ vectors in $$\mathbb{R}^d$$. Let me define the $$m \times m$$ matrix $$\mathbf{K}$$ such that its entries are given as \begin{align} \mathbf{K}_{ij} = \mathcal{k}(x_i,x_j) \end{align} For any vector $$x$$, let me define the $$m\times 1$$ vector $$\mathcal{K}_x$$ as \begin{align} \mathcal{K}_x = \begin{bmatrix} \mathcal{k}(x,x_1) \\ \vdots \\ \mathcal{k}(x,x_m) \end{bmatrix} \end{align} Now I am curious if following inequality holds true for every $$x\in\mathbb{R}^d$$ \begin{align} \mathcal{K}_x^T\mathbf{K}^{-1}\mathcal{K}_x \leq 1 \end{align}

The radial basis function kernel $$\mathcal{k}(x,y) = \exp\left(-\frac{\|x-y\|_2^2}{2\sigma^2}\right)$$ is a positive definite kernel. So, $$M=\begin{bmatrix}\textbf{K}&\mathcal{K}_x\\\mathcal{K}_x^T&1\end{bmatrix}$$ is a positive definie matrix. Hence, the Schur complement of the block $$1$$ of the matrix $$M$$ is also a positive definie matrix. That is $$1-\mathcal{K}_x^T\mathbf{K}^{-1}\mathcal{K}_x \geq 0$$. Q.E.D.
$$(\mathbf K^{-1}\mathcal{K}_x)_i = \sum_jK^{-1}_{ij}K_{jx}= \begin{cases} 1 & i=x\\ 0 & i \not = x \end{cases},$$
which is exactly $$e_x$$, the vector with a 1 in the $$x$$ coordinate and 0s elsewhere. We then remember that $$\mathcal{K}_x$$ is less than 1 in each component since $$e^{-y} \leq 1$$ for all $$y \in [0,\infty]$$ and then this gives us that $$\mathcal{K}_x^T\mathbf K^{-1}\mathcal{K}_x = \mathcal{K}_x^Te_x = k(x,x) = 1,$$
so the inner product is in fact 1. This generalizes to all invertible symmetric matrices with all diagonal elements $$\leq$$ 1.