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(-\frac{||x-y||_2^2}{2}) \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 $$\mathcal{k}(x,y) = \exp(-\frac{||x-y||_2^2}{2\sigma^2})$$ 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.