<s>[INST] If $a, b, c$ are positive and $a+b+c=1$, prove that $8abc\le\ (1-a)(1-b)(1-c)\le\frac{8}{27}$ If $a, b, c$ are positive and $a+b+c=1$, prove that $$8abc\le\ (1-a)(1-b)(1-c)\le\dfrac{8}{27}$$
I have solved $8abc\le\ (1-a)(1-b)(1-c)$ (by expanding $(1-a)(1-b)(1-c)$)
but do not get how to show that $(1-a)(1-b)(1-c)\le\frac{8}{27}$
 [/INST] For the 1st one: $(1-a)(1-b)(1-c) = (b+c)(a+c)(b+a) \ge 2\sqrt{bc}\cdot 2\sqrt{ac}\cdot 2\sqrt{ba} = 8abc$, and the 2nd one is by AM-GM as shown by the other answer.
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