23. Find the largest real number $m$ such that the inequality
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+m \leqslant \frac{1+x}{1+y}+\frac{1+y}{1+z}+\frac{1+z}{1+x}$$

holds for any positive real numbers $x, y, z$ satisfying $x y z=x+y+z+2$.
23. First guess the value of $m$ to be $\frac{3}{2}$, which is achieved if and only if $x=y=z=2$.

Let $x=\frac{1}{a}-1, y=\frac{1}{b}-1, z=\frac{1}{c}-1$, then we have $\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$, which means $a+b+c=1$. Thus, the original problem is to prove:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{3}{2} \leqslant \frac{b}{a}+\frac{c}{b}+\frac{a}{c}.$$

To prove inequality (11), it is sufficient to show:
$$\frac{a b}{c(b+c)}+\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)} \geqslant \frac{3}{2}.$$

Using the Cauchy-Schwarz inequality and the Arithmetic Mean-Geometric Mean inequality, it is easy to prove inequality (12). Indeed,
$$\begin{aligned}
& \frac{a b}{c(b+c)}+\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)} \\
= & \frac{(a b)^{2}}{a b c(b+c)}+\frac{(b c)^{2}}{a b c(c+a)}+\frac{(c a)^{2}}{a b c(a+b)} \\
\geqslant & \frac{(a b+b c+c a)^{2}}{2 a b c(a+b+c)} \\
= & \frac{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)}{2 a b c(a+b+c)} \\
\geqslant & \frac{a b c(a+b+c)+2 a b c(a+b+c)}{2 a b c(a+b+c)} \\
= & \frac{3}{2}
\end{aligned}$$

In conclusion, the maximum value of $m$ is $\frac{3}{2}$.