TITLE: Validity of $\sum\limits_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$?
QUESTION [10 upvotes]: Suppose that $\lambda_1\leq\lambda_2\leq\dots\leq\lambda_n$ is a sequence of real numbers. Clearly, if $a=(a_1,\dots, a_n)$ is a unit vector, then $\sum\limits_{i=1}^na_i^2\lambda_i\geq \lambda_1$. I want to see if the following generalization is true or not: 
If  $a=(a_1,\dots, a_n)$,  $b=(b_1,\dots, b_n)$, $c=(c_1,\dots, c_n)$, and $d=(d_1,\dots, d_n)$ ($n\geq 4$) form an orthonormal set, I wonder if we have
 $\sum\limits_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$.

REPLY [6 votes]: The proof of equality can also be handled economically via properties of the trace operator. Let $Q = (q_{ij})$ be an $n \times n$ orthogonal matrix and $\Lambda$ a diagonal matrix with entries $\lambda_1\leq \lambda_2 \leq \cdots \leq \lambda_n$. Then note that
$$
\newcommand{\tr}{\mathrm{tr}}
\sum_{i=1}^n \big(\sum_{j=1}^n q_{ij}^2\big) \lambda_i = \tr( Q^T \Lambda Q ) = \tr( Q Q^T \Lambda ) = \sum_{i=1}^n \lambda_i \>,
$$
and so we are done.
This can be generalized in the following way. Consider, instead, $Q$ as an $n \times m$ matrix with orthogonal columns (hence $m \leq n$). Then
$$
\sum_{i=1}^m \lambda_i \leq \sum_{i=1}^n \big(\sum_{j=1}^m q_{ij}^2\big) \lambda_i \leq \sum_{i=1}^m \lambda_{n-i} \>.
$$
This result can be deduced from the following theorem.
Theorem (Richter): Let $A$ and $B$ be $n \times n$ hermitian matrices with eigenvalues $\alpha_1 \leq \alpha_2 \leq \cdots \leq \alpha_n$ and $\beta_1 \leq \beta_2 \leq \cdots \leq \beta_n$, respectively. Then
$$
\sum_{i=1}^n \alpha_i \beta_{n-i} \leq \tr(AB) \leq \sum_{i=1}^n \alpha_i \beta_i \> .
$$
A simple proof of this result is found in

L. Mirsky, On the trace of matrix products, Math. Nachr., vol. 20, no. 3-6, pp. 171–174, 1959.

As pointed out in the aforementioned paper, there is a certain relationship with another more well-known result.
Theorem (von Neumann): Let $S$ and $T$ be arbitrary complex-valued $n \times n$ matrices. Let $(\sigma_i)$ and $(\tau_i)$ be the singular values of $S$ and $T$, respectively, in nonincreasing order. Then, $|\tr(ST)| \leq \sum_{i=1}^n \sigma_i \tau_i$.
A nice, elementary proof of this latter theorem can be found in

L. Mirsky, A trace inequality of John von Neumann, Monatsh. Math. 79 (4): 303–306, 1975, MR0371930.