Question

If $x=1+{\log }_{a}bc,y=1+{\log }_{b}ca,z=1+{\log }_{c}ab$, prove that $xy+yz+zx=xyz$

Answer 1

$x=1+{\log }_{a}bc$
$={\log }_{a}a+{\log }_{a}bc$
$x={\log }_a\left(abc\right)\Rightarrow \frac{1}{x}=\frac{1}{{\log }_a\left(abc\right)}={\log }_{abc}\left(a\right)$

Similarly,

$y={\log }_b\left(abc\right)\Rightarrow \frac{1}{y}={\log }_{abc}\left(b\right)$

$z={\log }_c\left(abc\right)\Rightarrow \frac{1}{z}={\log }_{abc}\left(c\right)$

Now, add all $3$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}={\log }_{abc}\left(a\right)+{\log }_{abc}\left(b\right)+{\log }_{abc}\left(c\right)$

$={\log }_{abc}\left(abc\right)$
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$
$xy+yz+zx=xyz$