Let me change the notations and reformulate the problem. You have a positive definite $n\times n$ ($n$ is your $K$) matrix $R$ with diagonal $D$ (your $D$ is $n$ times less than mine), and you have to prove that $nR^{-1}-D^{-1}$ is non-negative definite. Denote $R=D^{1/2}QD^{1/2}$, then $Q$ is a non-negative matrix with 1's along diagonal and we have to prove that $nR^{-1}-D^{-1}=D^{-1/2}(nQ^{-1}-I)D^{-1/2}$ is non-negative definite. Note that the sum of eigenvalues of $Q$ equals to the trace of $Q$, which equals to $n$. Therefore all eigenvalues of $Q$ belong to $(0,n)$, and all eigenvalues of $Q^{-1}$ belong to $(1/n,\infty)$, that just means that $nQ^{-1}-I$ is positive definite.

Let me change the notations and reformulate the problem. You have a positive definite $n\times n$ ($n$ is your $K$) matrix $R$ with diagonal $D$ (your $D$ is $n$ times less than mine), and you have to prove that $nR^{-1}-D^{-1}$ is non-negative definite. Denote $R=D^{1/2}QD^{1/2}$, then $Q=D^{-1/2}RD^{-1/2}$ is a positive definite symmetric matrix with all diagonal elements equal to 1. And we have to prove that $nR^{-1}-D^{-1}=D^{-1/2}(nQ^{-1}-I)D^{-1/2}$ is non-negative definite. Note that the sum of eigenvalues of $Q$ equals to the trace of $Q$, which equals to $n$. Therefore all eigenvalues of $Q$ belong to $(0,n)$, and all eigenvalues of $Q^{-1}$ belong to $(1/n,\infty)$, that just means that $nQ^{-1}-I$ is positive definite.

Let me change the notations and reformulate the problem. You have a positive definite $n\times n$ ($n$ is your $K$) matrix $R$ with diagonal $D$ (your $D$ is $n$ times less than mine), and you have to prove that $nR^{-1}-D^{-1}$ is non-negative definite. Denote $R=D^{1/2}QD^{1/2}$, then $Q$ is a non-negative matrix with 1's along diagonal and we have to prove that $nR^{-1}-D^{-1}=D^{-1/2}(nQ^{-1}-I)D^{-1/2}$ is non-negative definite. Note that $Q$ the sum of eigenvalues of $Q$ equals to the trace of $Q$, which equals to $n$. Therefore all eigenvalues of $Q$ belong to $(0,n)$, and all eigenvalues of $Q^{-1}$ belong to $(1/n,\infty)$, that just yields that $nQ^{-1}-I$ is positive definite.

Let me change the notations and reformulate the problem. You have a positive definite $n\times n$ ($n$ is your $K$) matrix $R$ with diagonal $D$ (your $D$ is $n$ times less than mine), and you have to prove that $nR^{-1}-D^{-1}$ is non-negative definite. Denote $R=D^{1/2}QD^{1/2}$, then $Q$ is a non-negative matrix with 1's along diagonal and we have to prove that $nR^{-1}-D^{-1}=D^{-1/2}(nQ^{-1}-I)D^{-1/2}$ is non-negative definite. Note that the sum of eigenvalues of $Q$ equals to the trace of $Q$, which equals to $n$. Therefore all eigenvalues of $Q$ belong to $(0,n)$, and all eigenvalues of $Q^{-1}$ belong to $(1/n,\infty)$, that just means that $nQ^{-1}-I$ is positive definite.