Let $X\_1,\dots,X\_n$ denote i.i.d. standard Gaussian random variables. My question is: Do there exist absolute constants $C,c>0$ such that for every $\epsilon>0$ and positive real numbers $a\_1,\dots,a\_n>0$, we have the following bound


$\quad \Pr(\sum\_{i=1}^n a\_i X\_i^2 \le \epsilon \sum\_{i=1}^n a\_i)\le C\epsilon^c\quad ?$


Background: Results of Carbery and Wright (2001) give powerful anti-concentration inequalities for polynomials of Gaussian variables. If we replace the term $\epsilon \sum\_{i=1}^n a\_i$ by $\epsilon\sqrt{\sum\_{i=1}^n a\_i^2}$ the bound follows directly from Carbery-Wright with $c=1/2$. Here, however, we need a stronger bound in terms of the $\ell\_1$-norm rather than $\ell\_2$-norm of the polynomial.


The reason why I believe the claim is still true is that all the variables $X\_i$ appear squared so that there are no cancellations. Also, for $n=1$ the claim follows from simple Gaussian anti-concentration bounds with $c=1/2.$ 


Note: A simple application of Paley-Zygmund gives the bound $\Pr(\sum a\_i X\_i^2 \ge \epsilon\sum a\_i )\ge (1-\epsilon)^2/3.$ This, unfortunately does not imply the statement above. 



Anti-concentration of Gaussian quadratic form
We can show that
$$
\mathbb{P}\left(\sum\_ia\_iX\_i^2\le\epsilon\sum\_ia\_i\right)\le\sqrt{e\epsilon}
$$
so that the inequality holds with $c=1/2$ and $C=\sqrt{e}$.


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.
Without loss of generality, we can suppose that $\sum\_ia\_i=1$ (just to simplify the expressions a bit). Then, for any $\lambda\ge0$,
$$
\begin{align}
\mathbb{P}\left(\sum\_ia\_iX\_i^2\le\epsilon\right)&\le\mathbb{E}\left[e^{\lambda\left(\epsilon-\sum\_ia\_iX\_i^2\right)}\right]\cr
&=e^{\lambda\epsilon}\prod\_i\mathbb{E}\left[e^{-\lambda a\_iX\_i^2}\right]\cr
&=e^{\lambda\epsilon}\prod\_i\left(1+2\lambda a\_i\right)^{-1/2}\cr
&\le e^{\lambda\epsilon}\left(1+2\lambda\right)^{-1/2}.
\end{align}
$$
Take $\lambda=(\epsilon^{-1}-1)/2$ to obtain
$$
\mathbb{P}\left(\sum\_ia\_iX\_i^2\le\epsilon\right)\le e^{(1-\epsilon)/2}\sqrt{\epsilon}.
$$


