6. Let $a=\lg x+\lg \left[(y z)^{-1}-1\right], b=\lg x^{-1}+$ $\lg (x y z+1), c=\lg y+\lg \left[(x y z)^{-1}+1\right]$, and let the maximum of $a, b, c$ be $M$. Then the minimum value of $M$ is $\qquad$
(Fujian contribution)
6. Ig2.

From the known conditions, we have $a=\lg \left(x y^{-1}+z\right), b=\lg (y z+$ $\left.x^{-1}\right), c=\operatorname{ig}\left[(x z)^{-1}+y\right]$. 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 known conditions, $x, y, z$ are all positive numbers, 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 $x=y=z=1$, $u=2$, so the minimum value of $u$ is 2, hence the minimum value of $M$ is $\lg 2$.