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show that： $$\sum_{1\le i

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}

$$\sum_{1\le i

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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$n=p$ (e.g., by AM-GM). – 2017-02-14
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    Unless I made some error, your initial conjecture ($p=3$) is *wrong* for $4 \le n \le 9$ (choose $x_1=\ldots=x_n=1/n$). – 2017-02-14

1 Answers 1

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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

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    Now,the question,How find the maximum of the value $\sum_{1\le i – 2017-02-15