# On the Relationship between $\sum a_1a_2\ldots a_m$ and $\sum a_1$

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:

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

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

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

• What does the sum mean ? Do we sum over all $m$-tuples which are subsets of {$a_1,a_2,\cdots ,a_n$} ? Apr 5, 2022 at 8:51
• @Peter Cyclic sum. So for $n=3, m=2$, $\sum ab = ab + bc + ca$, etc. Apr 5, 2022 at 8:51
• See Maclaurin's inequality: $\left(\frac{1}{n}\sum_{k=1}^n a_k\right)^m \ge \frac{1}{\binom{n}{m}}\sum\limits_{1\le i_1 < i_2 < \cdots < i_m \le n} a_{i_1} a_{i_2}\cdots a_{i_m}$ Apr 5, 2022 at 10:42
• It seemed to be equivalent, but is $\displaystyle \sum_{1\le i_1 < i_2 < \ldots < i_m \le n}$ equivalent to cyclic sum? Apr 5, 2022 at 10:59
• 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 $n,m$. Apr 5, 2022 at 18:45