Intersection of "positive" open half-spaces Prove that the intersection of "positive" open half-spaces associated with any basis
$x_1,x_2, \ldots, x_n$ of a finite dimensional inner product space $(V,(\cdot,\cdot))$ is non-empty.
Recall that the "positive" open half-space associated with $x_i$ is the set $A_i=\{y \in V|(y,x_i)>0\}$.
I got a hint that: Let $z_i$ be the projection of $x_i$ on the orthogonal
complement of the subspace spanned by all basis vectors except $x_i$, and consider
$t =\sum r_iz_i $ when all $r_i > 0$.)
However I couldn't find the proof yet. Any help would be appreciated.
 A: I am assuming that $V$ is a real vector space.
Let $P$ be the intersection of the positive half spaces.
Let $\phi:\mathbb{R}^n \to V$ be given by $\phi(y) = \sum_k y_k x_k$, note that $\phi$ is a bijection.
Define an inner product on $\mathbb{R}^n$ by $\langle a,b \rangle_V = (\phi(a),\phi(b) )$, where $(\cdot, \cdot)$ is the inner product on $V$. Let 
$Q$ be the matrix defined by $[Q]_{ij} = \langle e_i ,e_j \rangle_V$.
We have $\langle a,b \rangle_V = \sum_{ij} a_i b_j [Q]_{ij} = \langle a , Qb \rangle$, where the latter inner product is the standard inner product on $\mathbb{R}^n$.
Let $A = \{ a | \langle a, e_k \rangle_V >0\}$, and note that
$\hat{a}= Q^{-T} (1,1,...,1)^T \in A$.
Since $P = \phi^{-1}(A)$, we see that $\phi^{-1}(\hat{a}) \in P$ and
hence $P$ is not empty.
A: $\newcommand{\on}{\operatorname}$
$\newcommand{\la}{\langle}$
$\newcommand{\ra}{\rangle}$
Here's a proof more along the lines of the hint prescribed. Let's let $\la \cdot,\cdot \ra$ denote the inner product for $V$ so we don't overburden parentheses.
Let's assign
$$U_i:=\on{span}( \{ \vec{x}_1,\dots, \vec{x}_{i-1}, \vec{x}_{i+1},\dots,\vec{x}_n\}).$$ 
Then we let $\vec{z}_i= \on{proj}_{U_i^\perp}(\vec{x}_i)$. 
Claim: the inner product $\la\vec{x}_i, \vec{z}_i\ra>0$. 
In general, for a subspace $U\subseteq V$ and vector $\vec{x}\in V$, we have the inequality 
$$\la \vec{x},\on{proj}_{U} (\vec{x}) \ra \geq 0,$$ because if we let $\{\vec{u}_1,\dots,\vec{u}_k\}$ be an orthonormal basis for $U$, then we can write
$$\la\vec{x},\on{proj}_U(\vec{x})\ra=\la\vec{x},\sum_{i=1}^k \la\vec{x},\vec{u}_i\ra \vec{u}_i\ra=\sum_{i=1}^k \la\vec{x},\vec{u}_i\ra^2\geq 0.$$
In the case above,the inequality must be strict because $\la\vec{x}_i, \on{proj}_{U_i^\perp} (\vec{x}_i) \ra =0$ would imply that $\vec{x}_i$ was in $U_i$, which would contradict the assumption that $\{ \vec{x}_1,\dots,\vec{x}_n\}$ is a basis. Thus the claim is proved.
Furthermore, $ \la \vec{x}_i, \vec{z}_j\ra=0$ when $i\neq j$, by construction. Thus, we can take $\vec{z}=\sum_{i=1}^n \vec{z}_n$ (or any other strict positive combination of $\vec{z}_i$'s) as a vector which satisfies the desired property that $\la \vec{z},\vec{x}_i \ra>0$ for all $i$.
