If x ∝ y and y ∝ z, then show that \frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy} ∝ \frac{1}{x} + \frac{1}{y} + \frac{1}{z}.
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}^2z^2 + {k_2}^2 z^2 + z^2\over {k_1}{k_2}z({k_2}z) + k_2 z (z) + z (k_1k_2z)= ({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
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