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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} P\left(\sum_ia_iX_i^2\le\epsilon\right)\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} e^{\lambda\epsilon}\left(1+2\lambda\right)^{-1/2}. \end{align} Take $\lambda=(\epsilon^{-1}-1)/2$ to obtain $$P\left(\sum_ia_iX_i^2\le\epsilon\right)\le \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\right)\le e^{(1-\epsilon)/2}\sqrt{\epsilon}.$$
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} 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 $$P\left(\sum_ia_iX_i^2\le\epsilon\right)\le e^{(1-\epsilon)/2}\sqrt{\epsilon}.$$