* 6. The sum of four positive numbers is 4, and the sum of their squares is 8. Then, the maximum value of the largest number among these four numbers is
$6.1+\sqrt{3}$.
Let $a \geqslant b \geqslant c \geqslant d>0$.
Given $a+b+c+d=4, a^{2}+b^{2}+c^{2}+d^{2}=8$.
Then
$$
\begin{array}{l}
b+c+d=4-a, \\
b^{2}+c^{2}+d^{2}=8-a^{2} .
\end{array}
$$

By the Cauchy-Schwarz inequality
$$
3\left(b^{2}+c^{2}+d^{2}\right) \geqslant(b+c+d)^{2},
$$

i.e., $3\left(8-a^{2}\right) \geqslant(4-a)^{2}$,
which is $a^{2}-2 a-2 \leqslant 0$.
Thus, $a \leqslant 1+\sqrt{3}$.
Therefore, the maximum value of $a$ is $1+\sqrt{3}$.
At this point, $b=c=d=1-\sqrt{3}$.