Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1/n$.

Consider for constant $t>0$, $$S_n=\sum_{i=1}^n X_i^2e^{-4\lambda_i t}$$

I can just apply the inequality in that question taking $\sqrt{n}X_i$ and $a_i=e^{-4\lambda_i t}$.

Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2e^{-4\lambda_i t} \le \frac{\epsilon}{n} \sum_{i=1}^n e^{-4\lambda_i t}|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=n^{\alpha}$ for $\alpha<0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2e^{-4\lambda_i t} \ge n^{\alpha-1}\sum_{i=1}^n e^{-4\lambda_i t}|\lambda_1,\dots,\lambda_n\right)\ge 1-\sqrt{en^{\alpha}}\quad. $$

Question: Can we show that $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2e^{-4\lambda_i t} \ge \frac{1}{n} \sum_{i=1}^n e^{-4\lambda_i t}\right)\ge 1-\sqrt{\frac{e}{n}} (\mbox{or something with high probability)}\quad ? $$