Inequality -- 2

The Cauchy-Schwarz Inequality gives $a^{-1} + b^{-1} + c^{-1} \; = \; \big(a^{-1} + b^{-1} + c^{-1}\big)\big(a + b + c\big) \; \ge \; \left(\frac{\sqrt{a}}{\sqrt{a}} + \frac{\sqrt{b}}{\sqrt{b}} + \frac{\sqrt{c}}{\sqrt{c}}\right)^2 \; = \; 9$ with equality when $a=b=c=\tfrac13$ . Thus $\begin{aligned} M & = \; \big(a^{-1}-1\big)\big(b^{-1}-1\big)\big(c^{-1}-1\big) \; =\ \frac{1}{abc} - \left(\frac{1}{ab} + \frac{1}{ac} + \frac{1}{bc}\right) + \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) - 1 \\ & = \; \frac{1}{a} + \frac{1}{b} - \frac{1}{c} - 1 \;\ge \; 9 - 1 \; = \; \boxed{8} \end{aligned}$ with equality achieved when $a=b=c=\tfrac13$ .

$M = (\frac 1a - 1)(\frac 1b - 1)(\frac 1c - 1)$ $\\$ $M = (\frac {1-a}{a})(\frac {1-b}{b})(\frac {1-c}{c})$ $\\$ Since $a+b+c=1$ . $\quad$ $1-a=b+c; 1-b = a+c; 1-c = a+b$ $\\$ $M = (\frac {b+c}{a})(\frac {a + c}{b})(\frac {a+b}{c})$ $\\$ $M \geq \frac {2\sqrt {bc} \cdot 2\sqrt{ac} \cdot 2\sqrt {ab}}{abc}$ [By AM-GM Inequality] $\\$ $M \geq \frac {8abc}{abc}$ $\\$ $M \geq 8$ $\\$ $M_{min} = 8$