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This problem is a generalization of my answer to this question:

How to prove $\sum_{i=1}^r (r-1)a_i^2\geq\sum_{i,j=1\\i\neq j}^r a_ja_i$?

Let $s_m =\sum\limits_{i_1, i_2, ..., i_m, \text{ all }i_l\text{ distinct}} \prod\limits_{l=1}^m a_{i_l} $.

Show that $s_3 =\sum\limits_{u, v, w \text{ all distinct}} a_ua_va_w \le (n-1)(n-2)\sum\limits_{k=1}^n a_k^3 $.

Conjecture: Show that $s_m \le (n-1)(n-2)...(n-m+1)\sum\limits_{k=1}^n a_k^m = \prod\limits_{j=1}^{m-1}(n-j)\sum\limits_{k=1}^n a_k^m =\dfrac{(n-1)!}{(n-m)!}\sum\limits_{k=1}^n a_k^m $.

This is true for $m=2$ (as I showed in the linked answer) and $m=3$ (this problem).

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Yes, it is true for arbitrary $m \ge 2$ and $a_1, \ldots ,a_n \ge 0$.

I'll start with an alternative proof for $m=2$: Using the AM-GM inequality we get $$ s_2 = \sum_{i \ne j} a_i a_j \le \sum_{i \ne j}\frac 12 \left( a_i^2 + a_j^2 \right) = \sum_{i \ne j} a_i^2 = \sum_{i=1}^n \sum_{\substack{j=1 \\ j \ne i}}^n a_i^2 = (n-1) \sum_{i=1}^n a_i^2. $$

This can be generalized to any integer $m \ge 2$: From the AM-GM inequality we get $$ s_m =\sum_{\substack{i_1, i_2, ..., i_m \\ \text{distinct}}} \left(\prod_{l=1}^m a_{i_l} \right) \le \sum_{\substack{i_1, i_2, ..., i_m \\ \text{distinct}}} \left(\frac 1m \sum_{l=1}^m a_{i_l}^m \right) = \frac 1m \sum_{l=1}^m \sum_{\substack{i_1, i_2, ..., i_m \\ \text{distinct}}} a_{i_l}^m $$ For symmetry reasons, the inner sum is independent of $l$. Therefore $$ s_m \le \sum_{\substack{i_1, i_2, ..., i_m \\ \text{distinct}}} a_{i_1}^m = (n-1)(n-2)\cdots(n-m+1) \sum_{i_1=1}^n a_{i_1}^m $$ The last equality holds because for each $i_1$ there are $(n-1)(n-2)\cdots(n-m+1)$ possible choices for $i_2, \ldots, i_m$.

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