Six, let $x, y, z, w \in \mathbf{R}$, and $x+2 y+3 z+4 w=1$. Find the minimum value of $s=x^{2}+y^{2}+z^{2}+w^{2}+(x+y+z+w)^{2}$.

(Weng Shiyou)
Six, notice that,
$$
1=2(x+y+z+w)-x+0 y+z+2 w \text {. }
$$

Then $1^{2} \leqslant\left[2^{2}+(-1)^{2}+0^{2}+1^{2}+2^{2}\right]$.
$$
\left[(x+y+z+w)^{2}+x^{2}+y^{2}+z^{2}+w^{2}\right] \text {. }
$$

Thus, $s \geqslant \frac{1^{2}}{2^{2}+(-1)^{2}+0^{2}+1^{2}+2^{2}}=\frac{1}{10}$.
When $y=0$, and $\frac{x+y+z+w}{2}=-x=z=\frac{w}{2}$, that is,
$$
x=-\frac{1}{10}, y=0, z=\frac{1}{10}, w=\frac{1}{5}
$$

the equality in (1) holds.
Therefore, the minimum value of $s$ is $\frac{1}{10}$.