If $\log_a(bc) = x$, $\log_b(ca) = y$, and $\log_c(ab) = z$, find the value of $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1}$.

Add 1 to $x$:

$$x + 1 = \log_a(bc) + \log_a a = \log_a(abc) \implies \frac{1}{x+1} = \log_{abc} a$$

Similarly:

$$y + 1 = \log_b(ca) + \log_b b = \log_b(abc) \implies \frac{1}{y+1} = \log_{abc} b$$

$$z + 1 = \log_c(ab) + \log_c c = \log_c(abc) \implies \frac{1}{z+1} = \log_{abc} c$$

Summing the reciprocals:

$$\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \log_{abc} a + \log_{abc} b + \log_{abc} c = \log_{abc}(abc) = 1$$

Answer: 1