Anti-concentration of Gaussian quadratic form 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. 
 A: I also would strongly suspect it holds with $c = 1/2$.  Here's an argument which, if I haven't made a mistake, gives the upper bound $O(\epsilon^{1/2} \ln^{1/2}(1/\epsilon))$. 
Assume without loss of generality that $a_1 \geq a_2 \geq \cdots \geq a_n$ and that $a_1 + \cdots + a_n = 1$.  Divide into two cases:
Case 1: $a_1 \geq \frac{1}{100\log(1/\epsilon)}$.  In this case it's sufficient to show that we already have $\Pr[a_1 X_1^2 \leq \epsilon] \leq O(\epsilon^{1/2} \ln^{1/2}(1/\epsilon))$. But this holds as you said because of the anticoncentration of a single Gaussian.
Case 2: $a_1 \leq \frac{1}{100\log(1/\epsilon)}$.  In this case, note that $\sigma^2 := \sum_i a_i^2 \leq a_1 \sum_i a_i \leq \frac{1}{100\ln(1/\epsilon)}$. Now the random variables $X_i^2$ are nice enough that one should be able to apply a Chernoff Bound to them.  (I think I could provide a source for this if necessary.) Since $Y := \sum a_i X_i^2$ has mean $1$ and standard deviation $\sqrt{2}\sigma$ (I think, maybe the constant $\sqrt{2}$ is wrong), we should have a statement like $\Pr[|Y - 1| \geq t \cdot \sqrt{2}\sigma] \leq \exp(-t^2/c)$, where $c$ is a quite modestly small universal constant.  So $\Pr[Y \leq \epsilon] \leq \Pr[|Y-1| \geq 1/2] \leq \exp(-\frac{1}{8c\sigma^2}) \leq \exp(-\frac{12.5}{c} \ln(1/\epsilon)) = \epsilon^{12.5/c}$, which is smaller than $\epsilon^{1/2}$ if $c$ is not too large, and even if $c$ is too large, one can make $100$ larger.
A: 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
$$
\mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\right)\le e^{(1-\epsilon)/2}\sqrt{\epsilon}.
$$
A: You can directly use Chernoff's inequality, to get $LHS \le e^{-\Lambda^*(\epsilon)}$, where $\Lambda^*(\epsilon) := (\epsilon-1-\log\epsilon)/2$ is the Fenchel-Legendre transform of the log-MGF $\Gamma$ of $X_1^2$ (this has been computed in this SE post). Simplifying then gives
$$
LHS \le e^{-\Lambda^*(\epsilon)} = e^{-(\epsilon-1-\log\epsilon)/2} = e^{(1-\epsilon)/2}\sqrt{\epsilon} \le \sqrt{e\epsilon}.
$$
A: It looks like all the previous answers (including mine) are quite off by a huge margin, if the vector $a=(a_1,\ldots,a_n)$ is somewhat dense in the sense that $\|a\|_1/\|a\|_\infty$ is substantially larger than $1$. For example, if $a_i=1/n$ for all $i$, then we would expect to have the upper-bound to be of orde $\epsilon^{\Omega(n)}$, and not $\epsilon^{1/2}$. See Fact 2.1 of this blogpost.
In this post, I will provide an upper-bound which has somewhat optimal dependence on $a$.

Main tool: concentration function under affine transformation
Given a random vector $Y$ taking values in $\mathbb R^m$, its concentration function is defined by setting, for any $\epsilon \ge 0$,
$$
\mathcal L(Y,\epsilon) := \sup_{y \in \mathbb R^m} \mathcal L_y(Y,\epsilon),\text{ where }\mathcal L_y(Y, \epsilon) := \mathbb P(\|Y-y\|_2 \le \epsilon). 
$$
Now, let $A$ be a deterministic $m \times n$ matrix and let $X=(X_1,\ldots,X_n)$ be a random vector taking values in $\mathbb R^n$, with independent components verifying
$$
\max_{1 \le i \le n}\mathcal L(X_i,\epsilon) \le \ell(\epsilon).
$$
It is well-known (e.g see Theorem 1.5 of this paper by M. Rudelson) that for any $\delta \in (0,1)$, there exists a positive constant $C_\delta$ such that
$$
\mathcal L(AX,\epsilon\|A\|_{HS}) \le (C_\delta \ell(\epsilon))^{(1-\delta)r(A)},
$$
where $r(A) := \|A\|_{HS}^2/\|A\|_{op}^2 \ge 1$. In particular, if the $X_i$'s have standard normal distribution, then we may take $\ell(\epsilon)=\epsilon$,
and get for any $\epsilon \ge 0$ and $\delta \in (0,1)$,
$$
\mathcal L(AX,\epsilon\|A\|_{HS}) \le (C_\delta \epsilon)^{(1-\delta)r(A)}.
$$
Application to OP's question
In particular, taking $A$ to be the $n \times n$ diagonal matrix $\mbox{diag}(\sqrt a)$ with $a = (a_1,\ldots,a_n)$, we have
$$
\begin{split}
\mathbb P(\sum_{i=1}^n a_i X_i^2 \le \epsilon \sum_{i=1}^n a_i) &= \mathbb P(\|AX\|_2 \le \sqrt\epsilon\|A\|_{HS})\\
& = \mathcal L_0(AX, \sqrt\epsilon\|A\|_{HS})\\
& \le \mathcal L(AX,\sqrt\epsilon\|A\|_{HS}) \\
&\le (C_\delta \sqrt \epsilon)^{(1-\delta)r(A)}\\
& = (C_\delta \sqrt{\epsilon})^{(1-\delta)s(a)},
\end{split}
$$
where $s(a) := \|a\|_1 / \|a\|_\infty \in [1,n]$ measures the sparsity of $a$. As a sanity check, if $a_i=c > 0$ for all $i \in [n]$, then $s(a) = \|a\|_1 / \|a\|_\infty = n$, and our upper-bound above recovers the correct order in $\epsilon$, namely $\epsilon^{\Omega(n)}$.
