Example 4 For all $a, b, c \in \mathbf{R}^{+}$, find the minimum value of $f(a, b, c)=\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}$.
Let's make the following substitutions: $x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}, z=\frac{c}{\sqrt{c^{2}+8 a b}}$,
then $x, y, z \in(0,+\infty) . x^{2}=\frac{a^{2}}{a^{2}+8 b c}$, hence, $\frac{1}{x^{2}}-1=\frac{8 b c}{a^{2}}$.
Similarly, we have $\frac{1}{y^{2}}-1=\frac{8 a c}{b^{2}}, \frac{1}{z^{2}}-1=\frac{8 a b}{c^{2}}$,
multiplying the above three equations, we get $\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right)\left(\frac{1}{z^{2}}-1\right)=512$.
Suppose $x+y+z\frac{\left[(x+y+z)^{2}-x^{2}\right]\left[(x+y+z)^{2}-y^{2}\right]\left[(x+y+z)^{2}-z^{2}\right]}{x^{2} y^{2} z^{2}} \\
=\frac{(y+z)(2 x+y+z)(x+z)(2 y+x+z)(x+y)(x+y+2 z)}{x^{2} y^{2} z^{2}} \\
\geqslant \frac{2 \sqrt{y z} \cdot 4 \sqrt[4]{x^{2} y z} \cdot 2 \sqrt{x z} \cdot 4 \sqrt[4]{y^{2} x z} \cdot 2 \sqrt{x y} \cdot 4 \sqrt[4]{x y z^{2}}=512, \text { contradiction }}{x^{2} y^{2} z^{2}}=4,
\end{array}
$$

Therefore, $x+y+z \geqslant 1$, so when $a=b=c$, $f(a, b, c)$ achieves its minimum value of 1.