Following this question https://mathoverflow.net/questions/95096/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 $t>0$, $$S_n=\sum_{i=1}^n X_i^2e^{-4\lambda_i t}$$


**Question: can we show that 
 Do there exist absolute constants $C,c>0$ such that for every $\epsilon>0$, we have the following bound
$$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2e^{-4\lambda_i t} \le \epsilon \sum_{i=1}^n e^{-4\lambda_i t}|\lambda_1,\dots,\lambda_n\right)\le C\epsilon^c\quad ?$$**

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I just copied the proof in the first answer but he assume that $\sum_i a_i=1$ which is not true in my case.

>For $\epsilon\ge1$ the right hand side is greater than 1, so the inequality is trivial. I'll prove the case with $\epsilon < 1$ now. Here $a_i=e^{-4\lambda_i t}$.
Then, for any $\lambda\ge0$,
$$
\begin{align}
\mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\right)&\le\mathbb{E}\left[e^{\lambda\left(\epsilon-\sum_ia_iX_i^2\right)}\right]\cr
&=e^{\lambda\epsilon}\prod_i\mathbb{E}\left[e^{-\lambda a_iX_i^2}\right]\cr
&=e^{\lambda\epsilon}\prod_i\mathbb{E}\left[e^{-\lambda e^{-4\lambda_i t} X_i^2}\right]
\end{align}
$$