Let $g,\hat{g}$ be two Riemannian metrics with volume forms $dv\_g$, $dv\_{\hat{g}}$, respectively. A standard estimate in the subject is the following: $$\text{tr}\_g(\hat{g}) \leq \text{tr}\_{\hat{g}} (g)^{n-1}\frac{dv\_{\hat{g}}}{dv\_g},$$ where $n$ is the dimension.


In particular, if $g$ and $\hat{g}$ are related by some Monge-Ampere equation, one can often control $\text{tr}\_{g}(\hat{g})$ by $\text{tr}\_{\hat{g}}(g)$.


**Question:** Has anyone seen alternatives to this inequality? That is, suppose one has control of the ratio of volume forms, and one of the traces, is there another way of getting an estimate on the other trace?


To make the question more precise, fix a point $p$ such that $g(p) = \delta(p)$ (Euclidean metric) and $\hat{g}(p) = \lambda\_i \delta\_{ij}(p)$. Then the displayed inequality reads: $$\text{tr}\_g(\hat{g}) = \sum\_i \lambda\_i \leq \left( \sum\_i \lambda\_i^{-1} \right)^{n-1} \prod\_i \lambda\_i,$$ which appears to be some formulation of the arithmetic-geometric mean inequality.



Alternative to well-known trace estimate in Riemannian geometry?
There appear to be no alternatives, following the answer given by River Li over on MSE to [the more pedestrian formulation of this question](https://math.stackexchange.com/questions/4203635/powers-of-am-gm-hm-triples/4203956#4203956).


For posterity, let me give the details here: Let $g,\hat{g}$ be two Riemannian metrics. The estimate is local, so fix a point $p$ and choose coordinates such that $g$ is Euclidean at $p$ and $\hat{g}$ is diagonal with eigenvalues $\lambda\_i$. The original estimate $$\text{tr}\_g(\hat{g}) \leq \text{tr}\_{\hat{g}}(g)^{n-1} \frac{dv\_{\hat{g}}}{dv\_g}$$ is then written $$\sum\_i \lambda\_i \leq \left( \sum\_i \lambda\_i^{-1} \right)^{n-1} \prod\_i \lambda\_i.$$


Let $\text{AM} := \frac{1}{n} \sum\_i \lambda\_i$ denote the arithmetic mean, $\text{GM} := \left( \prod\_i \lambda\_i \right)^{\frac{1}{n}}$ denote the geometric mean, and $\text{HM} := n \left( \sum\_i \lambda\_i^{-1} \right)^{-1}$ denote the harmonic mean. Expresed in terms of these means, the aforementioned inequality is $$n \text{AM} \leq \frac{1}{n^{n-1}} \text{HM}^{1-n} \text{GM}^n,$$
or equivalently, $$\text{AM} \cdot \text{HM}^{n-1} \leq n^{-n} \text{GM}^n.$$


The original question, namely, whether there is an alternative to this inequality can therefore be more precisely formulated as, given constants $a,b,c, \lambda \in \mathbb{R}^+$ such that $$\text{AM}^a \text{HM}^b \leq \lambda \text{GM}^c,$$ are the only admissible candidates those which appeared in the original inequality?


In the answer to this [MSE question](https://math.stackexchange.com/questions/4203635/powers-of-am-gm-hm-triples/4203956#4203956), River Li gives us the following: Let $n \geq 2$ be an integer. Then $$\text{AM}^a \cdot \text{HM}^b \leq \lambda \text{GM}^{a+b}$$ if and only if $a/b \geq n-1$.


Note that the formulation in the MSE question has the inequality reversed, but is equivalent (take $\hat{g}$ to be Euclidean, and take $g$ to be diagonal).


If one spells out the condition $\frac{a}{b} \geq n-1$, one recovers that the only admissible powers are those in the original inequality.


