8. Given $a, b, c \in \mathbf{R}_{+}, a^{2}+b^{2}+c^{2}=1$. Let
$$
M=\max \left\{a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}\right\},
$$

then the minimum value of $M$ is . $\qquad$
8. $\frac{4 \sqrt{3}}{3}$.

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geqslant 3^{2}=9 \\
\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant 9 .
\end{array}
$$

From the problem, we know
$$
M \geqslant a+\frac{1}{b}, M \geqslant b+\frac{1}{c}, M \geqslant c+\frac{1}{a} .
$$

Then $3 M^{2}$
$$
\begin{array}{l}
\geqslant\left(a^{2}+b^{2}+c^{2}\right)+\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)+2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) \\
\geqslant 1+9+2 \times 3=16 \\
\Rightarrow M \geqslant \frac{4 \sqrt{3}}{3},
\end{array}
$$

with equality holding if and only if $a=b=c=\frac{\sqrt{3}}{3}$. Thus, the minimum value of $M$ is $\frac{4 \sqrt{3}}{3}$.