TITLE: Powers of AM-GM-HM triples
QUESTION [4 upvotes]: Let AM, GM, HM respectively denote the arithmetic, geometric and harmonic mean. An elementary consequence of Maclaurin's inequality shows that $$\text{AM}^{n-1} \text{HM} \geq \text{GM}^n.$$
Relating to the following problem I raised on MO, concerning an alternative to a well-known trace estimate in Riemannian Geometry, I'm an interested in whether an inequality of the form $$\text{AM}^a \text{HM}^b \geq \lambda \text{GM}^c,$$ for $a,b,c, \lambda >0$ forces $a = n-1, b = 1$ and $c=n$? Here, $\lambda$ is a negligible numerical constant.
Of course, the means all scale homogeneously of degree $1$, and therefore, an elementary scaling argument forces $a+b = c$. Moreover, taking various powers of the original inequality should be deemed equivalent.

REPLY [1 votes]: Some thoughts
Fact 1:  Let $n \ge 2$ be a positive integer. There exist positive real constants $a, b, \lambda$ such that
$\mathrm{AM}^a \mathrm{HM}^b \ge \lambda \mathrm{GM}^{a + b}$ holds for all positive reals $x_1, x_2, \cdots, x_n$ if and only if $a/b \ge n - 1$.

Proof of Fact 1:
“if” part: The inequality is written as
$\left(\frac{\mathrm{AM}}{\mathrm{GM}}\right)^{a/b} \ge \lambda^{1/b}\frac{\mathrm{GM}}{\mathrm{HM}}$.
Since $\mathrm{AM}^{n - 1} \mathrm{HM} \ge \mathrm{GM}^n$
which is written as $(\frac{\mathrm{AM}}{\mathrm{GM}})^{n - 1} \ge \frac{\mathrm{GM}}{\mathrm{HM}}$, noting that $\frac{\mathrm{AM}}{\mathrm{GM}} \ge 1$, we have $\left(\frac{\mathrm{AM}}{\mathrm{GM}}\right)^{a/b} \ge (\frac{\mathrm{AM}}{\mathrm{GM}})^{n - 1}\ge \frac{\mathrm{GM}}{\mathrm{HM}}$, and the desired result follows.
“only if” part: Let $x_1 = x_2 = \cdots = x_{n - 1} = x$ and $x_n = 1$.
From $\frac{\mathrm{AM}^a \mathrm{HM}^b}{\mathrm{GM}^{a + b}} \ge \lambda$,
we have $\lim_{x \to \infty} \frac{\mathrm{AM}^a \mathrm{HM}^b}{\mathrm{GM}^{a + b}} \ge \lambda$
which results in $a/b \ge n - 1$ (easy).
We are done.
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