10. (This question is worth 20 points) Let $a, b, c$ all be greater than 1, and satisfy
$$
\left\{\begin{array}{l}
\lg a+\log _{b} c=3, \\
\lg b+\log _{a} c=4 .
\end{array}\right.
$$

Find the maximum value of $\lg a \cdot \lg c$.
Solution: Let $\lg a=x, \lg b=y, \lg c=z$, from $a, b, c>1$ we know $x, y, z>0$.
From the conditions and the change of base formula, we have $x+\frac{z}{y}=3, y+\frac{z}{x}=4$, which means $x y+z=3 y=4 x$.
$\qquad$
Thus, let $x=3 t, y=4 t(t>0)$, then $z=4 x-x y=12 t-12 t^{2}$. From $z>0$ we know $t \in(0,1)$. $\qquad$
Therefore, combining the three-variable mean inequality, we get
$$
\begin{array}{l}
\lg a \lg c=x z=3 t \cdot 12 t(1-t)=18 \cdot t^{2}(2-2 t) \\
\leq 18 \cdot\left(\frac{t+t+(2-2 t)}{3}\right)^{3}=18 \cdot\left(\frac{2}{3}\right)^{3}=\frac{16}{3} .
\end{array}
$$

When $t=2-2 t$, i.e., $t=\frac{2}{3}$ (the corresponding $a, b, c$ are $100,10^{\frac{8}{3}}, 10^{\frac{8}{3}}$), $\lg a \lg c$ reaches its maximum value $\frac{16}{3}$. 20 points