10. [60] On a board the following six vectors are written:
$$
(1,0,0), \quad(-1,0,0), \quad(0,1,0), \quad(0,-1,0), \quad(0,0,1), \quad(0,0,-1) .
$$

Given two vectors $v$ and $w$ on the board, a move consists of erasing $v$ and $w$ and replacing them with $\frac{1}{\sqrt{2}}(v+w)$ and $\frac{1}{\sqrt{2}}(v-w)$. After some number of moves, the sum of the six vectors on the board is $u$. Find, with proof, the maximum possible length of $u$.
Answer: $2 \sqrt{3}$
Solution: For a construction, note that one can change
$$
(1,0,0),(-1,0,0) \rightarrow(\sqrt{2}, 0,0),(0,0,0) \rightarrow(1,0,0),(1,0,0)
$$
and similarly for $(0,1,0),(0,-1,0)$ and $(0,0,1),(0,0,-1)$. Then $u=(2,2,2)$.
For the bound, argue as follows: let the vectors be $v_{1}, \ldots, v_{6}, n=(x, y, z)$ be any unit vector, and $S=\sum_{i}\left(n \cdot v_{i}\right)^{2}$, where the sum is over all vectors on the board. We claim that $S$ is invariant. Indeed, we have
$$
\begin{aligned}
\left(n \cdot \frac{1}{\sqrt{2}}(v+w)\right)^{2}+\left(n \cdot \frac{1}{\sqrt{2}}(v-w)\right)^{2} & =\left(\frac{n \cdot v+n \cdot w}{\sqrt{2}}\right)^{2}+\left(\frac{n \cdot v-n \cdot w}{\sqrt{2}}\right)^{2} \\
& =\frac{2(n \cdot v)^{2}+2(n \cdot w)^{2}}{2} \\
& =(n \cdot v)^{2}+(n \cdot w)^{2}
\end{aligned}
$$

Also, at the beginning we have $S=2 x^{2}+2 y^{2}+2 z^{2}=2$. Therefore we must always have $S=2$.
Thus, by the Cauchy-Schwarz inequality we have
$$
n \cdot u=\sum n \cdot v_{i} \leq \sqrt{\sum_{i}\left(n \cdot v_{i}\right)^{2}} \sqrt{6}=\sqrt{12}=2 \sqrt{3} \text {. }
$$

But since $n$ is arbitrary, this implies that $|u| \leq 2 \sqrt{3}$; otherwise we could pick $n=u /|u|$ and reach a contradiction.