If \( x=\log _{a} b c, y=\log _{b} c a \) and \( z=\log _{c} a b \) then \( \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}= \)
\(1 + x = \log_a a + \log_a bc = \log_a(abc)\), so \(\frac{1}{1 + x} = \log_{abc}a\). Similarly for y and z.
Sum = \(\log_{abc}a + \log_{abc}b + \log_{abc}c = \log_{abc}(abc) = 1\)
Answer: (C) 1
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