# Sharpness of Arithmetic-Geometric inequality for eigenvalues

Let $A$ a $n\times n$ symmetric positive-definite matrix of eigenvalues $\lambda_n \leq \ldots \leq \lambda_1$. Then we have the estimate: $$\left(\frac{n-1}{\textrm{tr } A}\right)^{n-1} \textrm{det } A \leq \lambda_n,$$ by applying the arithmetic-geometric mean inequality on the set $\{\lambda_1,\ldots, \lambda_{n-1}\}$ of cardinality $n-1$. This is very coarse in general, but for a diagonal matrix $\ell \textrm{Id}$, it turns out to become $$\left(1-\frac{1}{n}\right)^{n-1} \ell \leq \ell,$$ so basically we are only off from a $e^{-1}$ factor.

Is there an example where this bound is sharp? If not is possible to get rid of this constant factor by tweaking things a bit?

The idea is the following. First try $n=2$ and you will find out the inequality is sharp and it becomes to be an equality if $\lambda_n=0$ (though $\lambda_n>0$ but we can take $\lambda\to 0^+$). Hence our first guess is that it is sharp when $\lambda_n=0$. Now we try to argue it.
Note that your inequality is equivalent to the following one $$\lambda_1\cdots \lambda_{n-1}\leqslant \left(\frac{\lambda_1+\cdots+\lambda_{n-1}+\lambda_{n}}{n-1}\right)^{n-1}$$ If we take $\lambda_n=0$, the inequality (just A-G inequality) is still true and the equality holds when $\lambda_1=\cdots=\lambda_{n-1}$. Now everything is clear. The inequality is sharp by taking $\lambda_1=\cdots=\lambda_{n-1}$ and $\lambda\to 0^+$.