<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] \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.
</s>