Suppose that a, b, and c are real numbers greater than 1. Then the value of \(\dfrac{1}{1+\log_{a^2 b} \frac{c}{a}} + \dfrac{1}{1+\log_{b^2 c} \frac{a}{b}} + \dfrac{1}{1+\log_{c^2 a} \frac{b}{c}}\) is

The sum is $3$. Since $1+\log_{a^2b}\frac ca=\log_{a^2b}(abc)$, each term equals $\log_{abc}(a^2b)$, so the sum is $\frac{\ln(a^2b)+\ln(b^2c)+\ln(c^2a)}{\ln(abc)}=\frac{3\ln(abc)}{\ln(abc)}=3$. Keeping only the squared factors in the numerators gives $2$, and keeping only the single factors gives $1$.