Following this question Anti-concentration of Gaussian quadratic form.

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

Moreover, based on the above inequality, taking $\epsilon=1/n$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2e^{-4\lambda_i t} \ge \frac{1}{n} \sum_{i=1}^n e^{-4\lambda_i t}|\lambda_1,\dots,\lambda_n\right)\ge 1-\sqrt{\frac{e}{n}}\quad $$ which means given $\lambda_1,\dots, \lambda_n$, then $$ \sum_{i=1}^n X_i^2e^{-4\lambda_i t} \ge \frac{1}{n} \sum_{i=1}^n e^{-4\lambda_i t} $$ with probability 1 as $n\to \infty$.

Question: How about we do not fix $\lambda_1,\dots, \lambda_n$? Do we still have $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2e^{-4\lambda_i t} \ge \frac{1}{n} \sum_{i=1}^n e^{-4\lambda_i t}\right)\ge 1-\sqrt{\frac{e}{n}}\quad ? $$