<s>[INST] How to prove the solution of these inequalities is empty? Prove: There does not exist 4 unit vectors $\mathbf{v}_1$, $\mathbf{v}_2$, $\mathbf{v}_3$, $\mathbf{v}_4$ in $\mathbb{R}^3$ such that
$$
\left \{ \ 
\begin{array}{ll}
\dfrac{4}{3} < \left \|\mathbf{v}_i + \mathbf{v}_j \right \| _2 ^2 < \dfrac{8}{3} \, , \\
\dfrac{4}{3} < \left \|\mathbf{v}_i - \mathbf{v}_j \right \| _2 ^2 < \dfrac{8}{3} \, ,
\end{array}
\ \  \forall\, 1\le i<j\le 4
\right.
.
$$
Note: It's preferred to prove it with algebra rather than high-level geometric concept.
 [/INST] Without loss of generality, take $$\mathbf{v}_1=(0,0,1)$$
Let $\displaystyle \mathbf{v}_{k}=\left( \sqrt{1-z_k^2}\cos \theta_{k},\sqrt{1-z_k^2}\sin \theta_{k},z_{k} \right)$ where $k=2,3,4$
$$|\mathbf{v}_1-\mathbf{v}_k|^2=2(1-z_k)$$
$$\frac{4}{3}< 2(1-z_k) < \frac{8}{3}$$
$$-\frac{1}{3} < z_{k} < \frac{1}{3}$$
To minimize $|\mathbf{v}_m-\mathbf{v}_n|$ for $2 \le m < n \le 4$, take $z_{k}=z$ and $\displaystyle \theta_{k}=\frac{2\pi k}{3}$
(We don't need to check for the case when $\theta_{m}=\theta_{n}$ since $|\mathbf{v}_{m}-\mathbf{v}_{n}|^2 < (\frac{2}{3})^2$ is smaller than the greatest lower bound.)
Now
$$|\mathbf{v}_m-\mathbf{v}_n|^2=3(1-z^2)$$
But $$\frac{8}{3} < 3(1-z^2) \le 3$$
That is $|\mathbf{v}_j-\mathbf{v}_k|$ is strictly larger than the least upper bound.
Contradiction.
</s>