TITLE: If $u_i\cdot u_j<0$ for all $u_0,...,u_n\in\Bbb R^n$, then the $u_i$ cannot lie in the same halfspace?
QUESTION [6 upvotes]: Given a set of $n+1$ vectors $u_0,...,u_n\in\Bbb R^n$ with pair-wise negative inner product, that is, $u_i\cdot u_j<0$ for $i\not=j$.

Question: what is a quick and clean way to see that the $u_i$ cannot all lie in the same half-space, that is, that there is no $y\in\Bbb R^n \setminus \{ 0 \}$ with $y\cdot u_i\ge 0$ for all $i\in\{0,...,n\}$?

REPLY [3 votes]: Assume that there exists such $y$. Use the idea of @Calvin Lin:   Denote $u_{n+1} = -y$. We have $u_i\cdot u_j < 0$ for $0\le i < j \le n$ and $u_i\cdot u_{n+1}\le 0$. 
We have $n+2$ vectors in $\mathbb{R}^n$. Add the component $1$ as the $n+1$-th component to each vector. Now we get $n+2$ vectors in $\mathbb{R}^{n+1}$, so linearly dependent. Conclude that there exists $a_0$, $\ldots$, $a_{n+1}$ not all $0$ so that
$$\sum_{i=0}^{n+1} a_i u_i=0$$ and $\sum a_i= 0$. Separate the numbers $a_i$ into positive and negative and ignore any zeros. We get an equality
$$\sum_{i\in I} p_i u_i = \sum_{j \in J} q_j u_j$$
for some non-void disjoint subsets $I$, $J$ and positive numbers $p_i$, $q_j$. 
From here we get 
$$0\le (\sum p_i u_i) \cdot (\sum p_i u_i) = \sum_{ij} p_i q_j\ u_i \cdot u_j \le 0$$
so all of the terms above are $0$. It follows that $I$ or $J$ consists only of $n+1$, but then $\|\sum p_i u_i\|^2>0$, contradiction.