At most $2n$ vectors, the angle between which $\geq\pi/2$. In a previous question it is proved that in $\mathbb R^n$ there are at most $n+1$ vectors, the angle between which $>π/2$.
How to prove that there are at most $2n$ vectors, the angle between which $\geq\pi/2$?
 A: Here is an elaboration of the comment by user1551. I will prove, by induction on the dimension:

If $E$ is a Euclidean vector space of dimension $n$, and $v_1,\ldots,v_k\in E\setminus\{0\}$ are such that $v_i\cdot v_j\leq0$ whenever $i\neq j$, then $k\leq 2n$.

Taking the vectors of an orthogonal basis and their opposites, one sees that the bound can be obtained.
Proof. The case $n=0$ is trivial, since $E\setminus\{0\}=\emptyset$ in that case. The case $n=1$ is also easy, and useful in the recursion: given one nonzero vector all other vectors must be negative multiples of it, but any two such multiples would have positive scalar product between them, so there can be at most two vectors in all. Now suppose that $n>1$ (and that $k>0$). Let $p:x\mapsto x-\frac{x\cdot v_1}{v_1\cdot v_1}v_1$ be the orthogonal projection onto $v_1^\perp$. Since the kernel  $\langle v_1\rangle$ of$~p$ has dimension$~1$, the case $n=1$ tells us that it it contains at most $2$ vectors$~v_i$ (among which of course $v_1$). So one gets at least $k-2$ nonzero vectors as projections in the $n-1$ dimensional Eucludean space $v_1^\perp$. We will be done by induction if we can show that the scalar products of the projections are nonpositive. This is easily checked: using that $p(x)\perp v_1$ one computes
$$
   v_i\cdot v_j = (p(v_i)+\frac{v_i\cdot v_1}{v_1\cdot v_1}v_1)\cdot(p(v_j)+\frac{v_i\cdot v_1}{v_1\cdot v_1}v_1) = p(v_i)\cdot p(v_j)+\frac{(v_i\cdot v_1)(v_j\cdot v_1)}{v_1\cdot v_1};
$$
since for $i\neq j$ one has $v_i\cdot v_j\leq0$ and $(v_i\cdot v_1)(v_j\cdot v_1)\geq0$ it follows that $p(v_i)\cdot p(v_j)\leq0$. $\qquad\square$
