y ∝ z ⇒ y = k₂z — (i)

x ∝ y ⇒ x = k₁y ⇒x = k₁k₂z — (ii)

\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy} \propto \frac{1}{x} + \frac{1}{y} + \frac{1}{z}

or, \frac{\text{x² + y² + z²}}{xyz} ∝ \frac{\text{yz + xz + xy}}{\text{xyz}}

or, \frac{\text{x² + y² + z²}}{\text{yz + xz + xy}} = k

To prove the proportionality for the given expression, the value of \frac{\text{x² + y² + z²}}{\text{yz + xz + xy}} should be non-zero constant.

Now substitute the values of x and y from (i) and (ii),

= ({k_1}^2{k_2}^2 + {k_2}^2  + 1)z^2\over {k_1}{k_2}^2 + k_2  + (k_1k_2)z^2

= ({k_1}^2{k_2}^2 + {k_2}^2  + 1)\over {k_1}{k_2}^2 + k_2  + (k_1k_2)

= Non – zero constant