If $x,y,z$ are positive and $x + y + z = 1$, then the minimum value of $\left( \frac{1}{x} - 1 \righ

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If $x,y,z$ are positive and $x + y + z = 1$, then the minimum value of $\left( \frac{1}{x} - 1 \right)\left( \frac{1}{y} - 1 \right)\left( \frac{1}{z} - 1 \right)$ is

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Sol. [Key Concept: $x + y + z = 1$: by AM–GM, $(\frac{1}{x} - 1)(\frac{1}{y} - 1)(\frac{1}{z} - 1)$ is minimised (value $8$) at $x = y = z = \frac{13}{}$.]$$\frac{(1 - x)(1 - y)(1 - z)}{xyz} = \frac{(y + z)(z + x)(x + y)}{xyz} \geq \frac{(2\sqrt{yz})(2\sqrt{zx})(2\sqrt{xy})}{xyz} = 8$$

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