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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 $(1-a)(1-b)(1-c)\le\frac{8}{27}$

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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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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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\begin{eqnarray*} a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0 \end{eqnarray*} This can be rearranged to \begin{eqnarray*} 9abc \leq (ab+bc+ca)(a+b+c) \\ 8abc \leq 1-(a+b+c)+(ab+bc+ca)-abc \end{eqnarray*} Thus the first inequality is shown.

\begin{eqnarray*} 4((a+b)(a-b)^2+(b+c)(b-c)^2+(c+a)(c-a)^2)+a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0 \end{eqnarray*} This can be rearranged to \begin{eqnarray*} 8(a^3+b^3+c^3)-3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)-6abc \leq 0 \\ 27(a+b+c)(ab+bc+ca)-27abc \leq 8(a+b+c)^3 \end{eqnarray*} Now add $27 -27(a+b+c)$ to the left hand side & use $a+b+c=1$. Thus the second inequality is shown.

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