Use Schur complement to show that $ 1^TD1 \ge 0$

Question

I'm trying to prove following statement:

Suppose $ D \preceq 0$ on the subspace $\{z: 1^Tz=0\}$. Use Schur complement to show that $1^TD1\ge0$.

I've done something like this

$1^TD1\ge0 \Leftrightarrow\left( \begin{matrix} 0 & 1^T \\ 1 & -D \end{matrix}\right) \succeq 0$

But I don't know what to do next. I would appreciate any advice.