If a = 1 + log_x yz, b = 1 + log_y zx and c = 1 + log_z xy, then show that

ab + bc + ca = abc.

a = 1 + log_x yz = log_x x + log_x yz = log_x xyz.

$\therefore \dfrac{1}{a} = \text{log}_{xyz} \space x$

b = 1 + log_y zx = log_y y + log_y zx = log_y xyz.

$\therefore \dfrac{1}{b} = \text{log}_{xyz} \space y$

c = 1 + log_z xy = log_z z + log_z xy = log_z xyz.

$\therefore \dfrac{1}{c} = \text{log}_{xyz} \space z$

$\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \text{log}${xyz} \space x + \text{log}{xyz} \space y + \text{log}_{xyz} \space z \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log x}}{\text{log xyz}} + \dfrac{\text{log y}}{\text{log xyz}} + \dfrac{\text{log z}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log x + log y + log z}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = \dfrac{\text{log xyz}}{\text{log xyz}} \\[1em] \Rightarrow \dfrac{bc + ac + ab}{abc} = 1 \\[1em] \Rightarrow ab + bc + ca = abc.

Hence, proved that ab + bc + ca = abc.