2015/044 ) If $x=1+log_a(bc)$ ,$y=1+log_b(ca)$ ,$z=1+log_c(ab)$ then, show that $xyz=xy+yz+zx$

we have $x = log_a a + log_a(bc) = log_a (abc)$

or $\dfrac{1}{x} = log_{abc} (a) \cdots  (1)$

$y = log_b b + log_b(ca) = log_b (abc)$

or $\dfrac{1}{y} = log_{abc} (b) \cdots(2)$

$z = log_c c + log_c(ab) = log_c (abc)$

or $\dfrac{1}{z} = log_{abc} (c) \cdots (3)$

add (1) (2) and (3) to get

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

or $\dfrac{1}{x}  +\dfrac{1}{y} + \dfrac{1}{z}=1$

multiplying both sides by xyz we get

$yz + zx + xy = xyz$


proved

this problem I picked from  https://in.answers.yahoo.com/question/index?qid=20150503005943AArBg7a