Let $\omega_1<\dots<\omega_n\in\mathbb{R}$. Then, define $G\in\mathbb{C}^{n\times n}$ such that $G_{k\ell}=\frac{1}{1-i(\omega_\ell-\omega_k)}$. For example, if $n=3$ we obtain $$ G=\begin{bmatrix}1 & \frac{1}{1-i(\omega_2-\omega_1)} & \frac{1}{1-i(\omega_3-\omega_1)}\\ \frac{1}{1+i(\omega_2-\omega_1)} & 1 & \frac{1}{1-i(\omega_3-\omega_2)}\\ \frac{1}{1+i(\omega_3-\omega_1)} & \frac{1}{1+i(\omega_3-\omega_2)} & 1 \end{bmatrix}$$ One can show that $G$ is positive definite. Now, let $\varepsilon<\omega_2-\omega_1$ and define $G_{\varepsilon}\in\mathbb{C}^{n\times n}$ just as $G$ but replacing $\omega_1$ by $\omega_1+\varepsilon$.
I want to prove $$\lambda_{\text{min}}(G_\varepsilon)\leq\lambda_{\text{min}}(G)$$ This is very easy to show in the $2\times 2$ case, as in that case we have a closed form formula for the least eigenvalue. Any comments on how could I prove it for the general $n\times n$ case? My thoughts to tackle the problem have been the following:
$\textit{1) Naive approach (failed)}$: it would suffice to show $G-G_\varepsilon\succeq 0$ and then conclude by Weyl's inequality. However, $G-G_\varepsilon\not\succeq 0$. This is because $\textrm{Tr}(G-G_\varepsilon)=0$, which implies $0=\sum_{k=1}^n\lambda_k(G-G_\varepsilon)$ and then that $G-G_\varepsilon$ has at least one negative eigenvalue (considering that it is Hermitian and non-zero).
$\textit{2) Variational approach (unclear)}$: let $h:\mathbb{R}_+\to\mathbb{R}$ such that $$ h(\varepsilon)=\lambda_{\textrm{min}}(G_\varepsilon). $$ By Weyl's inequality, $h$ is concave. Furthermore, the superdifferential of $h$ at $0$ can be characterized as $$ \partial h(0)=\{v^*G'v\,|\,v\in\mathbb{C}^n.\, Gv=\lambda_\textrm{min}(G)v,\, \|v\|=1\}\quad\text{ where}\quad G'=\lim_{\varepsilon\to 0}\frac{G_\varepsilon-G}{\varepsilon} $$ Then, if $v^*G'v\leq 0$ for some eigenvector $v$ associated to the smallest eigenvalue of $G$ we would have $$ \lambda_{\text{min}}(G_\varepsilon)=h(\varepsilon)\leq h(0)+\langle v^*G'v,\epsilon \rangle\leq h(0)=\lambda_{\text{min}}(G) $$ However, similarly as before, $\textrm{Tr}(G')=0$, and then $G'$ is not negative semi-definite. Then, for me its unclear how to prove that $v^*G'v\leq 0$ for some eigenvector $v$ associated to the smallest eigenvalue of $G$.
$\textit{Note}$: $G'$ has a structured form: $$ G'=\begin{bmatrix} 0 & x^*\\ x & 0 \end{bmatrix}\quad\text{where}\quad x\in\mathbb{C}^{n-1}.\ x_k=\frac{-1}{(1+i(\omega_{k+1}-\omega_1))^2} $$ Then, $\text{rank}(G')=2$, its non-zero eigenvalues are $\lambda_1=\|x\|_2$ and $\lambda_2=-\|x\|_2$, and they are associated to the eigenvectors $$ u_1=\begin{bmatrix} \|x\|\\ x \end{bmatrix}\quad\text{and}\quad u_2=\begin{bmatrix} -\|x\|\\ x \end{bmatrix} $$ respectively.
