Example 7 Let $a=\lg z+\lg \left[x(y z)^{-1}+1\right]$, $b=\lg x^{-1}+\lg (x y z+1), c=\lg y+$ $\lg \left[(x y z)^{-1}+1\right]$. Let $M$ be the maximum of $a$, $b$, and $c$. Then the minimum value of $M$ is $\qquad$
(1997, National High School Mathematics Competition)
Given the conditions, we have
$$
\begin{array}{l}
a=\lg \left(x y^{-1}+z\right), b=\lg \left(y z+x^{-1}\right), \\
c=\lg \left[(x z)^{-1}+y\right] .
\end{array}
$$

Let the maximum of \(x y^{-1}+z, y z+x^{-1}, (x z)^{-1}+y\) be \(u\), then \(M=\lg u\).
From the given conditions, \(x, y, z \in \mathbf{R}^{+}\), thus
$$
\begin{aligned}
u^{2} & \geqslant\left(x y^{-1}+z\right)\left[(x z)^{-1}+y\right] \\
& =\left[(y z)^{-1}+y z\right]+\left(x+x^{-1}\right) \\
& \geqslant 2+2=4 .
\end{aligned}
$$

Therefore, \(u \geqslant 2\), and when \(x=y=z=1\), \(u=2\). Hence, the minimum value of \(u\) is 2, and thus the minimum value of \(M\) is \(\lg 2\).