<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] Hint: $a,b,c \ge 0$ and $a+b+c=1$ implies $a, b, c \le 1\,$ so $1-a, 1-b, 1-c \ge 0$, then by AM-GM:
$$\sqrt[3]{(1-a)(1-b)(1-c)} \le \cfrac{(1-a)+(1-b)+(1-c)}{3}$$
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