Example 13 For all $a, b, c \in \mathbf{R}^{+}$, find the minimum value of $\frac{a}{\sqrt{a^{2}+8 b c}}$ $+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}}$.
Solution: Let the substitution be $x=\frac{a}{\sqrt{a^{2}+8 b c}}, y=\frac{b}{\sqrt{b^{2}+8 a c}}$, $z=\frac{c}{\sqrt{c^{2}+8 a b}}$, then $x, y, z \in(0,+\infty)$.

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,
we get $\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right)\left(\frac{1}{z^{2}}-1\right)=512$.
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[4]{x^{2} y^{3} z^{3}}\right)}{x^{2} y^{2} z^{2}}=512 . \text { Contradiction. }
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Therefore, $x+y+z \geqslant 1$. Hence, when $a=b=c$, the minimum value sought is 1.