Let $H$ be the hat/projection matrix that is symmetric and positive semidefinite and $D$ be a diagonal matrix with a diagonal of ones except the last term being greater than 1. If we partition these matrices in blocks we can say
$D= \begin{bmatrix}1&0&...&&\\0&1&0&... \\ & &1&... \\ \\ 0 &0&0&0&e>1\end{bmatrix} = \begin{bmatrix}\begin{bmatrix}I\end{bmatrix}&\begin{bmatrix}0\end{bmatrix}\\\begin{bmatrix}0\end{bmatrix}&e\end{bmatrix}$ and $ H = \begin{bmatrix}\begin{bmatrix}A\end{bmatrix}&\begin{bmatrix}C\end{bmatrix}\\\begin{bmatrix}C^T\end{bmatrix}&b\end{bmatrix}$ where C is a row vector and b is a scalar.
so
$ DH = \begin{bmatrix}\begin{bmatrix}A\end{bmatrix}&\begin{bmatrix}C\end{bmatrix}\\e\begin{bmatrix}C^T\end{bmatrix}&e\times b\end{bmatrix}$
The product DH will not be symmetrical but can we say anything about the truth of the following statement
$x^T DH x > 0$
where $x$ is a vector with all non-negative entries. I can't prove it's positive semidefinite from the beggining because the matrix is not symmetric. I thought maybe using Schur complements with these block matrices could work but I have not found any definitions for non symmetric matrices.
H is psd by definition. We can use the definitions of psd with the schur complement and state that $A\geq0$ and also $b - C^TA^{-1}C \geq 0$.
If we can use the same definition for the matrix DH(probably we can't because it's not symmetric) we know that $A>0$ and hence if $eb - eC^TA^{-1}C \geq 0$ the matrix is psd. Since e is a scalar bigger than 1 we see that the condition holds and we could proove that the matrix is psd.
Edit 1 - Edited to positive semidefinite
Edit 2 - Edited to have more information
Edit 3 - Proposal of a proof