On generalized AM-GM inequalities

Given $a_1,\dots,a_n \in \mathbb{R^+}$, and $2\le k\le n$, do we have:

$$\frac{\sum_\limits\text{cyclic}\sqrt[k]{a_1\dots a_k}}{\binom{n}{k}}\le \frac{\sum_\limits\text{cyclic}\sqrt[k-1]{a_1\dots a_{k-1}}}{\binom{n}{k-1}}$$

For example:

$$\dfrac{\sqrt[3]{a_1a_2a_3}}{\binom{3}{3}}\le\dfrac{\sqrt{a_1a_2}+\sqrt{a_1a_3}+\sqrt{a_2a_3}}{\binom{3}{2}}\le \dfrac{a_1+a_2+a_3}{\binom{3}{1}}$$

• That (almost) looks like en.wikipedia.org/wiki/Maclaurin%27s_inequality. – Martin R Nov 5 '15 at 13:09
• @Macavity It is in fact a valid inequality. – Quang Hoang Nov 5 '15 at 16:53
• @QuangHoang Yes, it's been edited to limit to positive reals. Haven't checked the rest, will do later – Macavity Nov 5 '15 at 16:59
• @Macavity I was talking about the second part of your comment. – Quang Hoang Nov 5 '15 at 17:01
• @QuangHoang That's what I said will do later. Not evident to me as trivial from the OP or from your answer. – Macavity Nov 5 '15 at 17:04

Yes, it is true and follows from usual AM-GM. For example \begin{aligned}&\sqrt[k]{a_2\cdots a_{k+1}}+\sqrt[k]{a_1a_3\cdots a_{k+1}}+\sqrt[k]{a_1a_2a_4\cdots a_{k+1}}+\cdots+\sqrt[k]{a_1a_2\cdots a_k}\\ &\ge (k+1)\cdot\sqrt[(k+1)k]{a_1^ka_2^k\cdots a_{k+1}^k}=(k+1)\cdot\sqrt[k+1]{a_1a_2\cdots a_{k+1}}.\end{aligned}\tag{1}

Note: the $m$-th term is $\sqrt[k]{\dfrac{a_1a_2\dots a_{k+1}}{a_m}}$.

The needed inequality follows by adding all similar inequalities.

Let's look at an example for $n=4,k=2$: \begin{aligned}\sqrt{a_2a_3}+\sqrt{a_1a_3}+\sqrt{a_1a_2}&\ge 3\sqrt[3]{a_1a_2a_3},\\ \sqrt{a_2a_4}+\sqrt{a_1a_4}+\sqrt{a_1a_2}&\ge 3\sqrt[3]{a_1a_2a_4}\\ \sqrt{a_3a_4}+\sqrt{a_1a_4}+\sqrt{a_1a_3}&\ge 3\sqrt[3]{a_1a_3a_4}\\ \sqrt{a_3a_4}+\sqrt{a_2a_4}+\sqrt{a_2a_3}&\ge 3\sqrt[3]{a_2a_3a_4} \end{aligned} Each of the $2$nd root occurs $n-k=4-2$ times (which I incorrectly counted as $n-k+1$ in the comment).

So how do we know for sure in general? There are two ways to see that

1. The precise way: For each $i_1<i_2<\cdots <i_k$, $\sqrt[k]{a_{i_1}a_{i_2}\cdots a_{i_k}}$ occurs exactly in the inequalities for $a_{i_1},a_{i_2},\cdots a_{i_k},a_{\color{red}{h}}$ for each $\color{red}{h} \neq i_1,i_2,\dots, i_k$.
2. The intuitive way: By symmetry of the inequality (which is kinda hard to explain), because we are talking about totally symmetry of (1), the means of $k$-th roots must be larger than or equal to the mean of $(k+1)$-th roots.

Big Edit I just realized that the OP asked for cyclic, not symmetric, sum. Shame on me. But that obviously is not a valid inequality when $a_i$ are all equal, just as @Macavity originally mentioned.

For example, when $n=4, k=2$ and all $a_i=1$, the RHS is $1$ while the LHS is $4/6$.

• For each $k+1$-tuple $a_{i_1},a_{i_2},\dots, a_{i_{k+1}}$ we have a similar inequality. Add them all up, the RHS gives the $(k+1)$ times of the $(k+1)$-th roots, while the LHS gives $n-k+1$ times of the $k$-th roots. – Quang Hoang Nov 5 '15 at 17:46
• The problem is stated for cyclic sums. I am not seeing those in your LHS, in fact counting all possible tuples would give you symmetric sums. Not so clear, perhaps you can illustrate for $n=4, k=3$ vs $n=4, k=2$ to show how your counting works. – Macavity Nov 5 '15 at 17:57
• In your first equation, the second term is $\sqrt[k]{a_1a_3...a_{k+1}}$. That term cannot appear in a cyclic sum. The same error is illustrated in your e.g. You may be proving something interesting, but it certainly is not the question posed by the OP, which is about cyclic sums. – Nemo Nov 5 '15 at 18:22
• @Nemo let me fix that. It is written not as a cyclic sum, but rather ordered sum. The case $k=2$ is a little mis-leading. – Quang Hoang Nov 5 '15 at 18:27
• Yes neither can $\sqrt{a_2a_4}$ appear in the cyclic sum. It's a symmetric sum you are putting down. Sorry this looks wrong as a proof. – Macavity Nov 5 '15 at 18:28