Consider pseudo-Euclidean space $\Bbb R^n_\nu$ with $\nu$ negatives first $(-,\ldots,-,+,\ldots,+)$. Split $\Bbb R^n_\nu = \Bbb R^\nu_\nu \times \Bbb R^{n-\nu}$. $\renewcommand\vec[1]{{\bf #1}}$Assume $\newcommand\pair[1]{\langle#1\rangle}$ that $\vec{u},\vec{v} \in \Bbb R^n_\nu$ are timelike vectors such that $$\vec{u} = \sum_{i=1}^\nu u^i \vec{e}_i + \vec{u}_0 \quad \mbox{and}\quad \vec{v} = \sum_{i=1}^\nu v^i \vec{e}_i + \vec{v}_0,$$with $u^i, v^i > 0$ for all $1 \leq i \leq \nu$, and $\vec{u}_0$ and $\vec{v}_0$ are spacelike vectors in $\{\vec{0}\} \times \Bbb R^{n-\nu}$, where $(\vec{e}_i)_{i=1}^n$ is the standard basis, say.
Conjecture: $\pair{\vec{u},\vec{v}} < 0$.
I know that for $\nu = 1$ this is true. Here's my attempt: denote by $\pair{\cdot,\cdot}_0$ the usual product in $\Bbb R^\nu$, and $\widetilde{\vec{u}} = \sum_{i=1}^\nu u^i \vec{e}_i$, and similarly for $\widetilde{\vec{v}}$.
Since $\vec{u}$ is timelike, we have $$0>\pair{\vec{u},\vec{u}} = -\sum_{i=1}^\nu (u^i)^2 + \pair{\vec{u}_0,\vec{u}_0} \implies \pair{\vec{u}_0,\vec{u}_0} < \pair{\widetilde{\vec{u}},\widetilde{\vec{u}}}_0,$$hence $\|\vec{u}_0\| \leq \|\widetilde{\vec{u}}\|_0$ and $\|\vec{v}_0\| \leq \|\widetilde{\vec{v}}\|_0$. Then, applying Cauchy-Schwarz (the usual one), we have $$\pair{\vec{u},\vec{v}} = - \pair{\widetilde{\vec{u}},\widetilde{\vec{v}}}_0 + \pair{\vec{u}_0,\vec{v}_0} \leq - \pair{\widetilde{\vec{u}},\widetilde{\vec{v}}}_0 + \|\vec{u}_0\|\|\vec{v}_0\| < - \pair{\widetilde{\vec{u}},\widetilde{\vec{v}}}_0 + \|\widetilde{\vec{u}}\|_0\|\widetilde{\vec{v}}\|_0, $$but this last quantity is always positive (again, Cauchy-Schwarz). I need to sharpen this bound somehow. Help?
The conjecture is false. Take for example $n=3$ and $\nu =2$. We can take $u = (2,0.1,1)$ and $v=(0.1,2,1)$. We have $\langle u,u \rangle <0$, $\langle v,v \rangle <0$ and $\langle u,v \rangle = 1-0.4>0$.