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show that： $$\sum_{1\le i<j<k\le n}x_{i}x_{j}x_{k}(x_{i}+x_{j}+x_{k})\le\dfrac{1}{27}?$$

over all $ n -$tuples $ (x_1, \ldots, x_n),$ satisfying $ x_i \geq 0$ and $ \sum_{i=1}^{n} x_i =1.$

I conjecture:let $p\in N^{+}$ $$\sum_{1\le m_{1}<m_{2}<\cdots<m_{p}\le n}x_{m_{1}}x_{m_{2}}\cdots x_{m_{p}}(x_{m_{1}}+x_{m_{2}}+\cdots+x_{m_{p}})\le\dfrac{1}{p^p}?$$ I tried C-S, but without success.

$$\sum_{1\le i<j<x_{k}\le n}x_{i}x_{j}x_{k}(x_{i}+x_{j}+x_{k})=\dfrac{1}{6}\sum_{i,j,k=1}^{n}x^2_{i}x_{j}x_{k}-\sum_{i=1}^{n}x^4_{i}??$$

math110
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    I haven't completely thought about this, but here is how I would approach the problem. Suppose that $x_1\geq x_2\geq \ldots \geq x_n$ and let $l$ be such that $l$ is largest and $x_l>0$. If $l>p$, then the sequence $\left(y_1,y_2,\ldots,y_n\right)$ given by $y_i:=x_i$ for all $i=1,2,\ldots,l-2$, $y_{l-1}:=x_{l-1}+x_l$, and $y_j=0$ for $j\geq l$ satisfies $$\sum_{1\leq m_1<\ldots<m_p\leq n}\sum_{i=1}^px_{m_i},\prod_{j=1}^px_{m_j}\leq \sum_{1\leq m_1<\ldots<m_p\leq n}\sum_{i=1}^py_{m_i},\prod_{j=1}^py_{m_j},.$$ So, if this works, you only need to prove when $n=p$ (e.g., by AM-GM). – Batominovski Feb 14 '17 at 09:06
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    Unless I made some error, your initial conjecture ($p=3$) is wrong for $4 \le n \le 9$ (choose $x_1=\ldots=x_n=1/n$). – Martin R Feb 14 '17 at 12:55

1 Answers1

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The inequality does not hold for $p \ge 3$ (at least not for arbitrary $n$). If you choose all $x_i$ equal to $\frac 1n$ then the left-hand side becomes $$ {n \choose p} \frac{p}{n^{p+1}} = {n-1 \choose p-1} \frac{1}{n^{p}} $$ For $n= p+1$ this is $$ \frac{p}{(p+1)^p} > \frac{1}{p^p} $$ where the last inequality follows from the fact that $x \to \sqrt[x]{x}$ is monotonically decreasing for $x > e$.

The counter example does not work for $p = 2 < e$, and in fact the inequality holds for $p=2$: If $\sum_{i=1}^{n}x_{i}=1$ and $x_{i}\ge 0$ then $\sum_{1\le i<j\le n}x_{i}x_{j}(x_{i}+x_{j})\le\frac{1}{4}$

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Martin R
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  • Now,the question,How find the maximum of the value $\sum_{1\le i<j<k\le n}x_{i}x_{j}x_{k}(x_{i}+x_{j}+x_{k})$ – math110 Feb 15 '17 at 13:09