On generalized AM-GM inequalities

Question

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:

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

Answer 1

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.

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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).

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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.

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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 $4/6$.