Symmetric tensor satisfying the dominant energy condition has $T^{00} \geq |T^{\alpha\beta}|$. In Hawking & Ellis's Large-Scale Structure of Spacetime, I've come across the following claim: If $T_{ab}$ is a symmetric tensor on a Lorentzian manifold $(M, g)$ satisfying the dominant energy condition, i.e. for all $W$ timelike vectors, 1) $T(W, W) \geq 0$, and 2) the vector $T^{ab}W_b$ is not spacelike, then
$$
T^{00} \geq |T^{\alpha\beta}|
$$
in any orthonormal basis.
I am confused on how to show this. Some partial progress: Let $\{e_0, e_1, e_2, e_3\}$ be an orthonormal basis with $e_0$ timelike. Setting $W = e_0$ and using property (2) should give
$$
|T^{00}|^2 \geq \sum_{i = 1}^3 |T^{0i}|^2
$$
which proves the claim in the case of $\alpha = 0$. However, I cannot show the claim for $T^{\alpha\beta}$ in general. My idea was to select $W = e_0 + ce_i$ for $i = 1, 2, 3$, with $c \in (0, 1)$ (so $W$ is timelike) and then use (2). But the most I got was showing the inequality
$$
\sum_{j = 1}^3 (|T^{0j}| - c |T^{ij}|)^2 \leq (T^{00} - c T^{0i})^2.
$$
I have tried expanding and using Cauchy-Schwarz on the cross-terms, but then I get wrong signs on things.  It's worth noting that this does prove that $T^{00} = 0 \implies$ all components of $T$ are zero, which is really all I need, but I would like to know a proof of the full claim from their book.
 A: Let's write
\begin{align}
W &= (u_0,\mathbf{u}),  \\
T &= 
  \begin{pmatrix}
    t_0        & \mathbf{t}^T \\
    \mathbf{t} & \mathbf{T} 
  \end{pmatrix},
\end{align}
then
\begin{align}
W\cdot W   &= u_0^2-|\mathbf{u}|^2>0, \\
TW         &= (u_0t_0+\mathbf{u}\cdot\mathbf{t},\ 
               u_0\mathbf{t}+\mathbf{T}\mathbf{u}), \\
T(W,W)     &= u_0^2t_0-\mathbf{u}\cdot\mathbf{T}\mathbf{u}\geq0, \\
TW\cdot TW &= (u_0t_0+\mathbf{u}\cdot\mathbf{t})^2-
              |u_0\mathbf{t}+\mathbf{T}\mathbf{u}|^2\geq0
\end{align}
Let's take $\mathbf{u}=\mathbf{0},$ then
\begin{align}
W\cdot W   &= u_0^2>0, \\
T(W,W)     &= u_0^2t_0\geq0, \\
TW\cdot TW &= u_0^2(t_0^2-|\mathbf{t}|^2)\geq0
\end{align}
from which we deduce
$$
t_0\geq0,\qquad t_0\geq|\mathbf{t}|\geq|t_i|,\quad\forall i=1,2,3.
$$
Now, let's take $u_0=1$ and $\mathbf{u}$ such that $|\mathbf{u}|<1,$ then
$$
(t_0+\mathbf{u}\cdot\mathbf{t})^2\geq|\mathbf{t}+\mathbf{T}\mathbf{u}|^2
$$
and if we change sign to $\mathbf{u}$ we have
$$
(t_0-\mathbf{u}\cdot\mathbf{t})^2\geq|\mathbf{t}-\mathbf{T}\mathbf{u}|^2
$$
If we add together these two last inequalities, we have
$$
t_0^2+(\mathbf{u}\cdot\mathbf{t})^2\geq
|\mathbf{t}|^2+|\mathbf{T}\mathbf{u}|^2\tag1
$$
Taking into account that
$$
(\mathbf{u}\cdot\mathbf{t})^2\leq|\mathbf{u}|^2|\mathbf{t}|^2<|\mathbf{t}|^2
$$
we can rewtite $(1)$ as
$$
t_0^2+|\mathbf{t}|^2>|\mathbf{t}|^2+|\mathbf{T}\mathbf{u}|^2
$$
and simplifying
$$
t_0^2>|\mathbf{T}\mathbf{u}|^2
$$
If we choose $\mathbf{u}=(\alpha,0,0),$ with $0<\alpha<1,$ we get
$$
t_0^2>\alpha^2[T_{11}^2+T_{21}^2+T_{31}^2]\geq\alpha^2T_{i1}^2
$$
and for $\alpha\to1^-$ we get the desired result
$$
t_0\geq|T_{i1}|,\quad\forall i=1,2,3.
$$
Choosing $\mathbf{u}=(0,\alpha,0)$ and $\mathbf{u}=(0,0,\alpha),$ we can get the other inequalities as well.
