2016/039) if $x=log_abc, y= log_bca, z= log_cab$ then show that $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 1$

we have $x=log_abc$ hence $x + 1 = log_abc + log_aa = log_aabc$ or $\frac{1}{x+1} = log_{abc}a$
 similarly $\frac{1}{y+1} = log_{abc}b$ and $\frac{1}{z+1} = log_{abc}c$
 Hence $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = log_{abc}a + log_{abc}b = log_{abc}c = log_{abc}abc =1$