Substitution Could Work?

$\begin{aligned} \frac 1{x+1} + \frac 1{y+1} + \frac 1{z+1} & = \frac 1{\log_a bc+1} + \frac 1{\log_b ca +1} + \frac 1{\log_c ab+1} \\ & = \frac 1{\frac {\log bc}{\log a}+1} + \frac 1{\frac {\log ca}{\log b}+1} + \frac 1{\frac {\log ab}{\log c}+1} \\ & = \frac {\log a}{\log b + \log c + \log a} + \frac {\log b}{\log c + \log a + \log b} + \frac {\log c}{\log a + \log b + \log c} \\ & = \boxed 1 \end{aligned}$

First note: $\log_{a}bc+1=\log_{a}abc$ etc so the substitution can yield the expression

$\frac{1}{\log_{a}abc}+\frac{1}{\log_{b}abc}+\frac{1}{\log_{b}abc}$

Next, use the change of base formula to make all base $a$ and simplify

$\frac{1}{\log_{a}abc}+\frac{\log_{a}b}{\log_{a}abc}+\frac{\log_{a}c}{\log_{a}abc}=\frac{\log_{a}a+\log_{a}b+\log_{a}c}{\log_{a}abc}=\frac{\log_{a}abc}{\log_{a}abc}=\boxed{1}$