TITLE: Prove the inequality: $\prod_{j=1}^ka_{jj}\leq\left(\frac{1}{k}\sum_{j=1}^k\lambda_j\right)^k.$
QUESTION [16 upvotes]: To all those who are eagerly awaiting a new question, all those who love math, I give this challenge and I hope for you good moments of reflection.
Let $A=(a_{ij})_n$ a real nonnegative symmetric matrix with eigenvalues $\lambda_1\geq\cdots\geq \lambda_n\geq 0$.
How to prove that, for all $k\in\{1,\ldots,n\}$, we have
$$\prod_{j=1}^ka_{jj}\leq\left(\frac{1}{k}\sum_{j=1}^k\lambda_j\right)^k.$$

REPLY [3 votes]: These inequalities hold more generally for $A\in M_n(\mathbb{C})$ hermitian semidefinite positive. My favourite argument uses the most convenient min-max theorem, which says that when the eigenvalues of $A$ are in nonincreasing order $\lambda_1\geq \lambda_2\geq\ldots \geq \lambda_n\geq 0$, we have
$$
\lambda_j=\max_{\dim F=j}\;\;\min_{x\in F, \|x\|=1} (Ax,x)
$$
where the max is taken over all dimension $j$ subspaces $F$ of $\mathbb{C}^n$.
Denoting $\{e_1,\ldots,e_n\}$ the canonical basis of $\mathbb{C}^n$, we see that $a_{jj}=(e_j,Ae_j)\geq 0$ for every $j$. By AM-GM applied to $a_{11},\ldots,a_{kk}$, we see that it suffices to prove that
$$
\sum_{j=1}^k a_{jj}\leq \sum_{j=1}^k\lambda_j \qquad\forall 1\leq k\leq n.
$$
In the case $k=n$, we easily see that this is an equality as both numbers are equal to the trace of $A$. It is more delicate to handle the case $k\leq n-1$.
Now we denote $A_k$ the upper-left square block of $A$ of size $k$, i.e. $A_k=(a_{ij})_{1\leq i,j\leq k}$. Note that it is a hermitian positive semidefinite matrix in $M_k(\mathbb{C})$ as $(A_kx,x)=(Ax,x)$ for every $x\in \mathbb{C}^k$ identified with the subspace of $\mathbb{C}^n$ spanned by $e_1,\ldots,e_k$. Writing $\mu_1\geq\ldots\geq \mu_k\geq 0$ the eigenvalues of $A_k$, the min-max theorem applied to $A_k$ yields
$$
\mu_j=\max_{F\subseteq \mathbb{C}^k\dim F=j}\;\;\min_{x\in F, \|x\|=1} (A_kx,x)=\max_{F\subseteq \mathbb{C}^k\dim F=j}\;\;\min_{x\in F, \|x\|=1} (Ax,x).
$$
Since the maximum of the rhs runs over a subset of the set of all dimension $j$ subspaces of $\mathbb{C}^n$, it is not greater than the maximum over the latter. Therefore
$$
\forall 1\leq j\leq k\qquad
\mu_j\leq \lambda_j\quad\Rightarrow\quad \sum_{j=1}^ka_{jj}=\mbox{tr}A_k=\sum_{j=1}^k\mu_j\leq \sum_{j=1}^k\lambda_j.
$$