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I am just starting learning inequalities and came across this inequality:

Given $n, m \in \mathbb{N^*}$, for any $a_1, a_2, \ldots, a_n \in \mathbb{R^+}$, is there a relationship between $\sum{a_1a_2\ldots a_m}$ and $(\sum a_1)^m$?

I can work out some of the special cases with AM-GM, for example:

$n =3, m = 2\implies\sum ab \le \dfrac{1}3(\sum a)^2$,

$n =4, m = 2\implies\sum ab \le \dfrac{1}4(\sum a)^2$,

$n =4, m = 3\implies\sum abc \le \dfrac{1}{16}(\sum a)^3$

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  • $\begingroup$ What does the sum mean ? Do we sum over all $m$-tuples which are subsets of {$a_1,a_2,\cdots ,a_n$} ? $\endgroup$
    – Peter
    Commented Apr 5, 2022 at 8:51
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    $\begingroup$ @Peter Cyclic sum. So for $n=3, m=2$, $\sum ab = ab + bc + ca$, etc. $\endgroup$
    – Lily White
    Commented Apr 5, 2022 at 8:51
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    $\begingroup$ See Maclaurin's inequality: $\left(\frac{1}{n}\sum_{k=1}^n a_k\right)^m \ge \frac{1}{\binom{n}{m}}\sum\limits_{1\le i_1 < i_2 < \cdots < i_m \le n} a_{i_1} a_{i_2}\cdots a_{i_m}$ $\endgroup$
    – achille hui
    Commented Apr 5, 2022 at 10:42
  • $\begingroup$ It seemed to be equivalent, but is $\displaystyle \sum_{1\le i_1 < i_2 < \ldots < i_m \le n}$ equivalent to cyclic sum? $\endgroup$
    – Lily White
    Commented Apr 5, 2022 at 10:59
  • $\begingroup$ Hmm... I didn't notice you are talking about cyclic sum. this is the sum over all combination and isn't equivalent to cyclic sum for general $n,m$. $\endgroup$
    – achille hui
    Commented Apr 5, 2022 at 18:45

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