Linear independency of vectors in Euclidean space I'm having trouble finishing my proof for the following problem:

Let $V$ be a Euclidean vector space. Let $n\geq 1, v_1, ..., v_n, w \in V$ with the following properties:
  $$\forall 1\leq i,j\leq n, i \neq j: \langle v_i, w \rangle > 0 \text{ and } \langle v_i, v_j \rangle \leq 0.$$
  Show that $v_1, ... v_n$ are linearly independent.

Proof:
Suppose $\sum_{i=1}^n \lambda_iv_i = 0$ for $\lambda_i \in \mathbb R$.
We have $$0 = \langle \sum_{i=1}^n \lambda_iv_i, w\rangle = \sum_{i=1}^n \lambda_i \langle v_i, w \rangle$$
(I don't know if I get information out of this. I at least know that $\lambda_i$ can't be all positive since $\langle v_i, w \rangle > 0$ for all $i$)
But also, for all $1\leq j \leq n$: $$0 = \langle \sum_{i=1}^n \lambda_iv_i, v_j \rangle = \sum_{i=1}^n \lambda_i \langle v_i, v_j \rangle = \sum_{i=1\\i\neq j}^n \lambda_i \langle v_i, v_j \rangle$$
So we in fact have one summand less than in the first equation. Does setting them equal help? Assumably I am nearly done, but I don't see how to conclude. Any help appreciated!
 A: First, it is important to interpret the given properties. While $\langle v_i,v_j\rangle\leq 0$ for $i\neq j$ refers to an angle of at least $90$ degree between each vectors $v_i$ and $v_j$ with $i\neq j$, you have $\langle v_i,w\rangle>0$ which yields an angle of less than $90$ degree between $v_i$ and $w$ for all $i$. If you try to imagine what the consequences are for $n=2$ or $n=3$ you get an idea for the proof.
The proof is seperated in three parts and maybe there is a shorter version...
The main idea is a proof by induction. If we consider $v_1,\ldots,v_{n+1}$ by induction hypothesis we get $v_1,\ldots,v_n$ linear independent. If $v_1,\ldots,v_{n+1}$ is not linear independent, we get $\lambda_1,\ldots,\lambda_n$ such that 
$$
v_{n+1}=\sum_{i=1}^n\lambda_iv_i.
$$
From the fact $\langle v_{n+1},v_i\rangle\leq 0$ for $1\leq i\leq n$ we like to conclude $\lambda_i\leq 0$ for $1\leq i\leq n$, such that we get a contradition to $\langle w,v_i\rangle>0$ for $1\leq i\leq n+1$. But the problem is, that $v_1,\ldots,v_n$ are not orthogonal. Therefore we use the trick to orthogonalize the vectors and check, that the properties still holds. This is the context of claim 1 and proposition 1. Using this idea yields the proof of your statement at the end.
Claim 1: Let be $n\geq 2$ and $v_1,\ldots,v_{n-1}$ linear independent. If $u_1\ldots,u_{n-1}$ is the corresponding set of orthogonalized vectors, using the Gram-Schmidt process, then $\langle v_n,u_i\rangle\leq 0$ and $\langle w,u_i\rangle>0$ for $1\leq i\leq n-1$.

Proof of claim 1: Prove by induction.
Base case $n=2$: Since $u_1:=v_1$ the statement is obvious.
Inductive step: Let be $n+1>2$ and $v_1,\ldots,v_n$ linear independent and $u_1,\ldots,u_n$ the corresponding orthogonalized vectors with the Gram-Schmidt process. By renaming the vectors and the induction hypothesis
we get $\langle v_{n+1},u_i\rangle\leq 0$ and $\langle w,u_i\rangle>0$ for $1\leq i\leq n-1$. We consider
$$
\langle v_{n+1}, u_n\rangle=\left\langle v_{n+1},v_n-\sum_{i=1}^{n-1}\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}u_i\right\rangle
=\langle v_{n+1},v_n\rangle-\sum_{i=1}^{n-1}\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}\langle v_{n+1},u_i\rangle.
$$
By induction hypothesis we have $\langle v_n,u_i\rangle\leq 0$ for $1\leq i\leq n-1$ and therefore the RHS is nonpositive and we get $\langle v_{n+1},u_n\rangle \leq 0$. Further we have
$$
\langle w,u_n\rangle=\left\langle w,v_n-\sum_{i=1}^{n-1}\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}u_i\right\rangle
=\langle w,v_n\rangle+\sum_{i=1}^{n-1}\frac{\left(-\langle v_n,u_i\rangle\right)}{\langle u_i,u_i\rangle}\langle w,u_i\rangle.
$$
which yields $\langle w,u_n\rangle>0$ the same way.
Proposition 1: If $v_1,\ldots,v_{n-1}$ are linear independent and $u_1,\ldots,u_{n-1}$ is the corresponding set of orthogonalized vectors, using Gram-Schmidt process and there exists $\mu_1,\ldots,\mu_{n-1}\in\mathbb{R}$ such that
$$
v_n=\sum_{i=1}^{n-1}\mu_iu_i,
$$
then $\mu_i\leq 0$ for $1\leq i\leq n-1$.
Proof of proposition 1:
From $v_n=\sum_{i=1}^{n-1}\mu_iu_i$ we get $\mu_i=\frac{\langle v_n,u_i\rangle}{\langle u_i,u_i\rangle}$, 
since $u_1,\ldots,u_{n-1}$ are orthogonal. Using claim 1 yields $\mu_i\leq 0$.
Proof of your statement: Prove by induction.
Base case
For $n=1$ there is nothing to proof, since $\langle v_1,w\rangle>0$ yields $v_1\neq 0$.
For $n=2$ from $0=\sum_{i=1}^2\lambda_i\langle v_i,w\rangle$ you get
$\lambda_1=0=\lambda_2$ and you are done or w.l.o.g. $\lambda_1<0<\lambda_2$. But $$0=\left\langle \sum_{i=1}^2 \lambda_iv_i, v_1 \right\rangle = \sum_{i=1}^2 \lambda_i \langle v_i, v_1 \rangle \Leftrightarrow -\lambda_1\langle v_1v_1\rangle= \lambda_2 \langle v_2, v_1 \rangle.
$$
The LHS is positive while the RHS is negative and we get a contradition. Therefore $\lambda_1=\lambda_2=0$ has to be true and $v_1,v_2$ are lineare independent.
Inductive step:
Now we consider $n+1>2$ and $v_1,\ldots,v_{n+1}\in V$. By induction hypothesis we get, that $v_1,\ldots,v_n$ are linear independent. Either $v_1,\ldots,v_{n+1}$ are lineare independent and we are done or we consider $u_1,\ldots, u_n$ the corresponding orthogonalized vectors by Gram-Schmidt process and there exists $\mu_1,\ldots,\mu_n$ such that
$$
v_{n+1}=\sum_{i=1}^n\mu_iu_i.
$$
We consider 
$$
0=\left\langle w , v_{n+1}-\sum_{i=1}^n\mu_i u_i\right\rangle=\langle w,v_{n+1}\rangle+\sum_{i=1}^n(-\mu_i)\langle w, u_i\rangle.
$$
By proposition 1 and claim 1 we have $(-\mu_i)\langle w,u_i\rangle\geq 0$ for $1\leq i\leq n$ which yields a contradition since $\langle w,v_{n+1}\rangle>0$. Therefore we get, that $v_1,\ldots,v_{n+1}$ are lineare independent.
A: We know that inner product can't be negative, and if is 0, the vectors are orthogonal. Orthogonal vectors are, by definition, linearly independent.
As for the w information, i think it was given to tell you that for any v vector, it is not the 0 vector, which is orthogonal to any vector (if it was, the inner product and v and w was 0).
