Prove the inequality: $\prod_{j=1}^ka_{jj}\leq\left(\frac{1}{k}\sum_{j=1}^k\lambda_j\right)^k.$ 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
$$\prod_{j=1}^ka_{jj}\leq\left(\frac{1}{k}\sum_{j=1}^k\lambda_j\right)^k.$$
 A: Apply G.M. $\le$ A.M. to $a_{jj}, j = 1,\ldots,k$, we have:
$$\prod_{j=1}^k a_{jj} \le \left( \frac{1}{k} \sum_{j=1}^k a_{jj} \right)^{k}$$
Since $A$ is a real non-negative symmetric matrix with non-negative eigenvalues $\lambda_1 \ge \lambda_2 \ge \cdots \lambda_n \ge 0$, we can find an orthogonal matrix $\Omega$ and diagonal matrix $\Lambda$ 
such that $A = \Omega^{T} \Lambda \Omega$ and $\Lambda_{ii} = \lambda_i$ for $i = 1,\ldots,n$. In terms of coefficients of $\Omega$, we have:
$$\sum_{j=1}^k a_{jj} = \sum_{j=1}^k \sum_{i=1}^n \lambda_i |\Omega_{ij}|^2
=\sum_{i=1}^n\lambda_i \beta_i
\,\,\,\text{ where }\,\,\,\beta_i = \sum_{j=1}^k |\Omega_{ij}|^2
$$
Using properties of orthogonal matrices, it is not hard to see $0 \le \beta_i \le 1$ and $\sum_{i=1}^n \beta_i = k$.
Let $x_0 = 0$ and $x_i = \sum_{s=1}^i \beta_s$ for $i = 1,\ldots,n$. Define 
two functions $\beta(t)$, $\gamma(t)$ on $[0,k)$ by:
$$\begin{align}
\beta(t)  &= \lambda_i \,\,\,\text{ for }\,\,\, t \in [x_{i-1}, x_i), & i = 1,\ldots, n\\
\gamma(t)  &= \lambda_i \,\,\,\text{ for }\,\,\, t \in [i-1, i), & i = 1,\ldots, k
\end{align}$$
Since $\lambda_i$ is non-increasing, it is easy to see $\beta(t) \le \gamma(t)$ on $[0,k)$. From this, we can deduce:
$$\sum_{i=1}^{n} \lambda_i\beta_i = \int_{0}^{k} \beta(t) dt \le \int_{0}^{k} \gamma(t) dt = \sum_{i=1}^k \lambda_i$$
As a result, we can conclude:
$$\prod_{j=1}^k a_{jj} \le \left( \frac{1}{k} \sum_{j=1}^k \lambda_j \right)^{k}$$
EDIT Background information.
The statement $\sum_{j=1}^k a_{jj} \le \sum_{j=1}^k \lambda_j$ is actually a corollary of a well known theorem first proved by Schur. Namely,

Let $A$ be an $n\times n$ Hermitian matrix. Let $\text{diag}(A)$ denote the vector     whose coordinates are the diagonal entries of $A$ and $\lambda(A)$ a vector whose coordinates are the eigenvalues of $A$ arranged in any order, then $\text{diag}(A)$ is majorized by $\lambda(A)$.
What this means is when we sort the components of $\text{diag}(A)$ and $\lambda(A)$ into two n-tuples of decreasing order:
  $$a^{\downarrow}_1 \ge a^{\downarrow}_2 \ge \cdots \ge a^{\downarrow}_n
\,\,\,\text{ and }\,\,\,
    \lambda^{\downarrow}_1 \ge \lambda^{\downarrow}_2 \ge \cdots \ge \lambda^{\downarrow}_n$$
  we have $a^{\downarrow}_1 \le \lambda^{\downarrow}_1$,
  $\,\,\,a^{\downarrow}_1 + a^{\downarrow}_2 \le \lambda^{\downarrow}_1 + \lambda^{\downarrow}_2$
  and in general,
  $$\sum_{i=1}^k a^{\downarrow}_i \le \sum_{i=1}^k \lambda^{\downarrow}_i
\,\,\,\text{ for } 1 \le k < n\,\,\,\text{ and }\,\,\,
\sum_{i=1}^n a^{\downarrow}_i = \sum_{i=1}^n \lambda^{\downarrow}_i$$

A: 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
$$
\forall 1\leq j\leq k\qquad
\mu_j\leq \lambda_j\quad\Rightarrow\quad \sum_{j=1}^ka_{jj}=\mbox{tr}A_k=\sum_{j=1}^k\mu_j\leq \sum_{j=1}^k\lambda_j.
$$
A: Hint: Suppose orthogonal matrix $P_k$ can diagonalize the matrix $A_{k}$, where $A_k$ means the matrix made by the first $k$ cols and $k$ rows of $A$.
As $P_kA_kP_k' = \Phi_k $ as a diagonal matrix, with diagonal as $\left[\mu_1,\mu_2,\cdots,\mu_k\right]$, arranged in decreasing order.
write $A$ as
$\left(\begin{array}[cc]
AA_k&U\\
U'&R
\end{array}\right)$
where $U$ as $k\times (n-k)$ matrix. $R$ as symmetric $(n-k)\times(n-k)$ matrix.
$\left(\begin{array}[cc]
AP_k&0\\
0&I_{n-k}
\end{array}\right)$$\left(\begin{array}[cc]
AA_k&U\\
U'&R
\end{array}\right)$
$\left(\begin{array}[cc]
AP_k'&0\\
0&I_{n-k}
\end{array}\right)$ = $\left(\begin{array}[cc]
A\Phi_k& P_kU\\
U'P_k'&R
\end{array}\right)$
And $tr(\Phi_k) = tr(A_k) = \sum_{i=1}^k a_{ii}$, take $\Lambda = \left[\lambda_1,\cdots,\lambda_n\right]$
If we can prove $\sum_{i=1}^k\lambda_i\ge tr(\Phi_k)$, then we can use AM-GM inequality to prove it.
take $x=(x_1,\cdots,x_k,0)$, $0$ is a vector in dimension $n-k$. 
Consider $z =x\left(\begin{array}[cc]
AP_k'&0\\
0&I_{n-k}
\end{array}\right) P'_n $,
then $\sum_{i=1}^n\lambda_i z_i^2 =\sum_{j=1}^k\mu_j x_j^2 $,
we take $x = \left[0,0,\cdots,1,0,0\cdots,0\right]$ for first $k$ basis.
when $x = e^m$, as $m$th basis, we got $z^m = (z_1^m,\cdots,z_n^m)$.
And it is easy to prove $z^l$,$z^m$ are orthogonal when $m\neq l$.
Therefore $\sum_{j=1}^k(z_i^j)^2\le 1$. [Note arrange all the $z^l$ as a matrix, and see this by looking at cols]
Since
$\sum_{i=1}^n \lambda_i (z_i^m)^2 = \mu_m$,
Thus
$\sum_{j=1}^k \mu_j = \sum_{i=1}^n\lambda_i(\sum_{j=1}^k(z_i^j)^2)\le \sum_{i=1}^k \lambda_i$.
Since
$\sum_{i=1}^n\sum_{j=1}^k(z_i^j)^2 = k$.
