110. Let $x=\lg a, y=\lg b, z=\lg c$, since $a, b, c$ are real numbers greater than 1, so $x, y, z >$ 0.
$$\begin{array}{l} \log _{a} b c+\log _{b} c a+\log _{c} a b \geqslant 4\left(\log _{a b} c+\log _{b c} a+\log _{c a} b\right) \Leftrightarrow \\ \frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y} \geqslant \frac{4 x}{y+z}+\frac{4 y}{z+x}+\frac{4 z}{x+y} \end{array}$$

It is only necessary to prove $\frac{z}{x}+\frac{z}{y} \geqslant \frac{4 z}{x+y}$, etc., which is equivalent to proving $\frac{1}{x}+\frac{1}{y} \geqslant \frac{4}{x+y}$. This is obvious.