Choose 3 vectors of length $\geq 1$ so that sums of each two have length $< 1$ Is it possible to choose three vectors $v_1, v_2, v_3\in\mathbb{R}^2,\vert v_i\vert\geq1$, so that
$$\vert v_1+v_2\vert<1$$
$$\vert v_2+v_3\vert<1$$
$$\vert v_3+v_1\vert<1$$
where $\vert v\vert$ is the euclidean norm of $v$?
I could not find a counterexample, but I also did not manage to prove it.
Since each $v_i$ only has two coordinates, I tried to prove it by distinction of all possible combinations of negative and positive coordinates for each vector, but this did not work out (and it would be a very clumsy proof if it worked).
I also tried a similar approach using polar coordinates, which also did not work.
Do you have an counterexample or an approach for a proof?
 A: Hint:   the three angles between pairs of the three vectors add up to at most $2\pi\,$, so at least one of them must be $\le 2 \pi /3\,$. Let WLOG that one be the angle $0 \le \alpha \le 2 \pi / 3$ between $v_1$ and $v_2\,$. Then $\cos(\alpha) \ge -1/2\,$, and therefore:
$$
\begin{align}
|v_1+v_2|^2 &= |v_1|^2+|v_2|^2+ 2 |v_1||v_2|\cos(\alpha) \\
 &\ge |v_1|^2+|v_2|^2 - |v_1||v_2| \\
 &= |v_1||v_2| + \big(|v_1|-|v_2|\big)^2 \\
 &\ge |v_1||v_2| \\
 &\ge 1
\end{align}
$$
A: For $i \neq j$ you have $|v_i+v_j|²<1$ which gives $|v_i|^2+|v_j|^2+2\langle v_i, v_j \rangle <1$ which means that $2\langle v_i, v_j \rangle<1-|v_i|^2-|v_j|^2<-1$.
Then write that there exist $(a,b,c) \neq (0,0,0)$ such that $av_1+bv_2+cv_3=0$ since you have 3 vectors in a space of dimension 2.
You can ensure that at least 2 of the 3 coefs $a,b,c$ are non-negative (y multiplying all of them by $-1$ if it is not the case)
If one of them is negative, say $c$ then $-cv_3=av_1+bv_2$, compute the dot product with $v_3$: $-c|v_3|^2=a\langle v_1,v_3 \rangle + b\langle v_2,v_3 \rangle$, the lhs is positive while the rhs is negative which is absurd.
Then $a,b,c \geq 0$.
Compute the dot products by $v_1, v_2, v_3$ of the equality $av_1+bv_2+cv_3=0$:
$$a|v_1|^2+b\langle v_1, v_2 \rangle + c \langle v_1, v_3 \rangle=0$$
$$b|v_2|^2+a\langle v_1, v_2 \rangle + c \langle v_2, v_3 \rangle=0$$
$$c|v_3|^2+a\langle v_1, v_3 \rangle + b \langle v_2, v_3 \rangle=0$$
and sum these equalities, you obtain:
$0 = a|v_1|^2+b|v_2|^2+c|v_3|^2+b\langle v_1, v_2 \rangle + c \langle v_1, v_3 \rangle+a\langle v_1, v_2 \rangle + c \langle v_2, v_3 \rangle+a\langle v_1, v_3 \rangle + b \langle v_2, v_3 \rangle$
And use the fact that $2\langle v_i, v_j \rangle<1-|v_i|^2-|v_j|^2$. You get (since $a,b,c \geq 0$):
$$0<a|v_1|^2+b|v_2|^2+c|v_3|^2 + (b/2)(1-|v_1|^2-|v_2|^2) + (c/2)(1-|v_1|^2-|v_3|^2)+(a/2)(1-|v_3|^2-|v_2|^2)+(c/2)(1-|v_2|^2-|v_3|^2)+(a/2)(1-|v_1|^2-|v_3|^2)+(b/2)(1-|v_2|^2-|v_3|^2)$$
i.e.,
$$a+b+c>(b/2)|v_1|^2+(c/2)|v_1|^2+(a/2)|v_2|^2+(c/2)|v_2|^2+(a/2)|v_3|^2+(b/2)|v_3|^2\geq a+b+c$$ which is absurd.
A: 
$$\begin{align}
2\alpha+2\beta+2\gamma = 360^\circ \quad&\stackrel{\text{wlog}}{\implies}\quad \alpha \geq 60^\circ \\[4pt]
&\implies\quad \left(\;\alpha\geq\beta^\prime\;\right)\;\text{or}\;\left(\;\alpha\geq\gamma^\prime\;\right) \\[4pt]
&\implies\quad \left(\;a^\prime \geq b\;\right)\;\text{or}\;\left(\;a^\prime \geq c\;\right) \quad\square
\end{align}$$
