Show that $s_3 =\sum_{u, v, w \text{ all distinct}} a_ua_va_w \le (n-1)(n-2)\sum_{k=1}^n a_k^3 $. Generalize. 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).
 A: 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 $i_2, \ldots, i_m$.
