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Prove that (1-a)(1-b)(1-c) >= 8abc if a+b+c=1

by RoRi

If a,b,c \in \mathbb{R}_{>0} and a+b+c = 1, prove that

    \[ (1-a)(1-b)(1-c) \geq 8abc. \]

Proof. Recall the definition of the pth power mean, M_p,

    \[ M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p} \]

for x_1, \ldots, x_n are n positive real numbers and p is any integer.
By the generalization of the Arithmetic Mean-Geometric Mean inequality (proved in part b of the linked exercise) we know that M_{-1} \leq M_1.

So, for the problem at hand, we consider M_{-1} (a,b,c) \leq M_1 (a,b,c) gives us,

    \[ \left( \frac{a^{-1} + b^{-1} + c^{-1}}{3} \right)^{-1} &\leq \left( \frac{a+b+c}{3} \right). \]

Simplifying this expression, we have

    \begin{align*} && \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} &\leq \frac{1}{3} & (a+b+c = 1)\\ \implies && 9 &\leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ \implies && 9abc &\leq bc + ac + ab & (\text{multiplying by } abc)\\ \implies && 8abc &\leq bc + ac + ab - abc \\ \implies && 8abc & \leq 1 - (a+b+c) + ab + ac + bc - abc \\ \intertext{where we are just adding 0 to the right hand side since $1 = a+b+c$ implies $1-(a+b+c) = 0$. Then since $1 - (a + b + c) + ab + ac + bc - abc = (1-a)(1-b)(1-c)$ (which we can check by multiplying out the right hand side), we have} \implies && 8abc &\leq (1-a)(1-b)(1-c). \qquad \blacksquare \end{align*}

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