6. Let $a=\lg z+\lg \left[x(y z)^{-1}+1\right], b=\lg x^{-1}+\lg (x y z+1), c=\lg y+\lg \left[(x y z)^{-1}+1\right]$, denote the maximum of $a, b$, $c$ as $M$, then the minimum value of $M$ is $\qquad$

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$\lg 2$
6.【Analysis and Solution】From the given conditions, we have $a=\lg \left(x y^{-1}+z\right), b=\lg \left(y z+x^{-1}\right), c$ $=\lg \left[(x y)^{-1}+y\right]$. Let the maximum number among $x y^{-1}+z, y z+x^{-1},(x z)^{-1}+y$ be $u$, then $M$ $=\lg u$.
From the given conditions, we know that $x, y, z$ are all positive numbers, thus
$$
\begin{array}{l}
u^{2} \geqslant\left(x y^{-1}+z\right)\left[(x z)^{-1}+y\right] \\
=\left[(y z)^{-1}+y z\right]+\left(x+x^{-1}\right) \geqslant 2+2=4 .
\end{array}
$$

Therefore, $u \geqslant 2$, and when $x=y=z=1$, $u=2$, so the minimum value of $u$ is 2, hence the minimum value of $M$ is $\lg 2$.