I came across the following inequality for the determinant:
$$ \frac{n}{\mathrm{tr}(A^{-1})}\leq \det (A)^{1/n}\leq \frac{1}{n}\mathrm{tr}(A)\leq \sqrt{\frac{1}{n}\mathrm{tr}(A^{2})} $$
which looks suspiciously like the HM-GM-AM-QM inequality:
$$ \frac{n}{a^{-1}+b^{-1}+\dots}\leq (a b\cdots)^{1/n}\leq \frac{a+b+\dots}{n}\leq \sqrt{\frac{a^2+b^2+\dots}{n}}. $$
Of course, the later is generalized to the generalized mean inequality:
$$\left( \frac{1}{n} \sum_{i=1}^n x_i^p \right)^{\frac{1}{p}} \le \left( \frac{1}{n} \sum_{i=1}^n x_i^q \right)^{\frac{1}{q}} \quad\text{for}\quad p \le q. $$
My question is whether the trace/determinant inequality generalizes in a similar way? such that:
$$ \left(\frac{1}{n}\mathrm{tr}(A^{p})\right)^{\frac{1}{p}} \le \left(\frac{1}{n}\mathrm{tr}(A^{q})\right)^{\frac{1}{q}} \quad\text{for}\quad p \le q \text{?} $$
This seems like something that would be true, but I wonder if it has a name, and if there is an intepretation of $\left(\frac{1}{n}\mathrm{tr}(A^{p})\right)^{1/p}$ as $p\to-\infty$ or $p\to\infty$, like there is for means?
Of course it is. You can show it by noting that $\operatorname{tr}(A^{p})$ is just the sum of the eigenvalues of $A^p$, and that the eigenvalues of $A^p$ are just the eigenvalues of $A$ raised to the power $p$, $\forall p\in\Bbb{N}$. But of course the eigenvalues need to be real.