Example 3 For all $a, b, c \in \mathbf{R}_{+}$, find
$$
\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}
$$

the minimum value.
Explanation: Make the substitution
$$
\begin{array}{l}
x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, \\
z=\frac{c}{\sqrt{c^{2}+8 a b}} .
\end{array}
$$

Then $x, y, z \in (0, +\infty)$.
Thus, $x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, which means $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, $\frac{1}{y^{2}}-1=\frac{8 a c}{b^{2}}, \frac{1}{z^{2}}-1=\frac{8 a b}{c^{2}}$.
Multiplying the above three equations yields
$$
\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right)\left(\frac{1}{z^{2}}-1\right)=512 \text{. }
$$

If $x+y+z\frac{\prod\left[\left(\sum x\right)^{2}-x^{2}\right]}{x^{2} y^{2} z^{2}} \\
=\frac{\prod[(y+z)(2 x+y+z)]}{x^{2} y^{2} z^{2}} \\
\geqslant \frac{\prod\left(2 \sqrt{y z} \cdot 4 \sqrt[4]{x^{2} y z}\right)}{x^{2} y^{2} z^{2}} \\
=\frac{\prod\left(8 \sqrt{x^{2} y^{3} z^{3}}\right)}{x^{2} y^{2} z^{2}}=512 .
\end{array}
$$

This leads to a contradiction.
Therefore, $x+y+z \geqslant 1$.
Thus, when $a=b=c$, the minimum value is 1.