8. If $a+b+c=1, a, b, c \in \mathbf{R}^{+}, \sqrt{4 a+1}+$ $\sqrt{4 b+1}+\sqrt{4 c+1}>m$, then the maximum value of $m$ is $\qquad$
8. $2+\sqrt{5}$.

Solution: First, prove that $2+\sqrt{5}$ is a lower bound.
Since $a b>0$, we have $(a+1)(b+1)>1+a+b$, which means $\sqrt{(a+1)(b+1)}>\sqrt{1+a+b}$,

Thus, $a+1+b+1+2 \sqrt{(a+1)(b+1)}>1+1+a+b+2 \sqrt{1+a+b}$,
which implies $(\sqrt{a+1}+\sqrt{b+1})^{2}>(1+\sqrt{1+a+b})^{2}$,
which further implies $\sqrt{a+1}+\sqrt{b+1}>1+\sqrt{1+a+b}$,
By repeatedly applying the above inequality, we get
$$
\begin{array}{l}
\quad \sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}>1+ \\
\sqrt{4(a+b)+1}+\sqrt{4 c+1} \\
>1+1+\sqrt{1+4(a+b+c)}=2+\sqrt{5},
\end{array}
$$

Additionally, when $a \rightarrow 1, b \rightarrow 0, c \rightarrow 0$, the left side of the inequality approaches $2+\sqrt{5}$, so $2+\sqrt{5}$ is the greatest lower bound.