Question: $A\in\mathbb{R}^{n\times n}$ is a positive definite matrix with constant trace, i.e., $A>0$ and $\mbox{tr} A=k$. Let $\lambda_1\ge\cdots\ge\lambda_n>0$ be the eigenvalues of $A$. Can we say maximizing $\det A=\prod_{i=1}^n \lambda_i$ is equivalent to minimizing $\mbox{tr} A^2=\sum_{i=1}^n \lambda_i^2$?
Remark: It seems a trivial problem. But maybe it is not so easy to solve. Since $\mbox{tr} A=\sum_{i=1}^n\lambda_i=k$, it is obvious that $\det A=\prod_{i=1}^n \lambda_i\le \left(\frac{\sum_{i=1}^n\lambda_i}{n}\right)^n=\left(\frac{k}{n}\right)^n$ and the maximum is achieved iff $\lambda_i=k/n, \forall i$. On the other hand, we have $\mbox{tr} A^2=\sum_{i=1}^n \lambda_i^2\ge\frac{(\sum_{i=1}^n\lambda_i)^2}{n}=\frac{k^2}{n}$ and the minimum is achieved iff $\lambda_i=k/n, \forall i$.
It seems $\det A$ and $\mbox{A}$ are equivalent because they reach optima simultaneously when $\lambda_i=k/n$. But in some cases $\lambda_i=k/n, \forall i$ is impossible, do they still reach optima simultaneously? For example, consider the constraints: for some $\lambda_i\le\alpha_i
My attempt: For a special case $n=2$, we have $\det A=((\mbox{tr}A)^2-\mbox{tr}(A^2))/2$. Since $\mbox{tr}A$ is constant, even if $\lambda_i=k/n, \forall i$ is not satisfied, we still can say $\mbox{tr}A^2$ and $\det A$ reach optimum at the same time. But for $n=3$ we have $\det A=((\mbox{tr}A)^3-3\mbox{tr}A\mbox{tr}A^2+\mbox{tr}A^3)/6$. Can we say $\det A$ and $\mbox{tr}A^2$ reach optima simultaneously? Thank you.