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Suppose $0\leq x_1\leq x_2\leq\ldots\leq x_{3n}$ are real numbers. We are to show that $$\left(\sum_{i=1}^{3n}x_i\right)^3\geq 27n^2\left(\sum_{i=1}^nx_ix_{n+i}x_{2n+i}\right).$$

Take $n=1$, this is $(x_1+x_2+x_3)^3\geq 27x_1x_2x_3$, which follows from the AM-GM inequality. Take $n=2$, this is $(x_1+\ldots+x_6)^3\geq 108(x_1x_3x_5+x_2x_4x_6)$. Maybe the assumption $x_1\leq\ldots\leq x_6$ is important here?

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  • $\begingroup$ Funny thing is that for $n\ge 2$ there's an equality when $x_1=x_2=\ldots=x_{n-1}=0$, $x_n=x_{n+1}=\ldots=x_{3n-1}=1$, and $x_{3n}=n$. So you have to be careful when making estimates to not break this equality. $\endgroup$ – user2345215 Nov 16 '14 at 19:43
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As the inequality is homogeneous, we can WLOG set $\displaystyle \sum_{k=1}^{3n} x_k = 3n$. With this condition, the inequality can be written as: $$n -\sum_{k=1}^n x_k x_{k+1}x_{2k+1} \ge 0$$

Now, the LHS can be considered a linear function of each $x_i$, with a non-positive coefficient. So the LHS can be minimised by setting each $x_i \in [x_{i-1}, x_{i+1}]$ to the maximum possible value allowed, i.e. $x_i = x_{i+1}$. Thus LHS is minimised when we have $ x_i = x_{3n}$ for each $i$, so $\displaystyle \sum_{k=1}^{3n} x_k =3n \implies \forall i, \;x_i=1$, and the minimum of the LHS is $\displaystyle n-\sum_{k=1}^n 1\cdot1\cdot1= n-n=0$. Hence the inequality holds true.

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