I'm dealing with the following chance constraint
$$ \mathbb{P}(\|Ax\|_2 \leq c^\intercal x) \geq 1 - \delta, $$
where $x \sim \mathcal{N}(\mu,\Sigma)$ and $\delta \in (0, 0.5]$. Since $x$ follows a normal distribution, we know that $\xi := \|Ax\|_2$ follows a $\chi$ distribution, so let's just say that has a p.d.f. $f_{\xi}(x)$. Similarly, $\eta := c^\intercal x$ will also a follow a normal distribution, i.e. $\eta \sim \mathcal{N}(c^\intercal \mu, c^\intercal \Sigma c)$, so let's say the p.d.f is $f_{\eta}(y)$ for simplicity. Then, the above equation boils down to
$$ \mathbb{P}(\xi \leq \eta) \geq 1 - \delta. $$
I am trying to prove that if $\mathbb{P}(\xi \leq \bar{\eta}) \geq 1 - \delta$ is satisfied, where $\bar{\eta} := \mathbb{E}[\eta] = c^\intercal \mu$, this implies that $\mathbb{P}(\xi \leq \eta) \geq 1 - \delta$. In order to prove this, I must prove that $\mathbb{P}(\xi \leq \eta) \geq \mathbb{P}(\xi \leq \bar{\eta})$. However, this is the part that I am stuck on.
My attempt is as follows: Suppose the joint probability distribution of the two random variables is given by $f_{\xi,\eta}(x,y)$. Then, by definition, the probability is given as
$$ \mathbb{P}(\xi \leq \eta) = \int_{-\infty}^{\infty}\int_{-\infty}^{y} f_{\xi,\eta}(x,y) \ \textrm{d}x \ \textrm{d} y. $$
Similarly, the the probability that $\xi$ is less than $\bar{\eta}$ is given by
$$ \mathbb{P}(\xi \leq \bar{\eta}) = \int_{-\infty}^{\bar{\eta}} f_{\xi}(x) \ \textrm{d} x. $$
Another thought is that the random variable $\eta$ can be broken down in terms of a standard normal distribution $z \sim \mathcal{N}(0,1)$ as $\eta = \bar{\eta} + z \sqrt{c^\intercal \Sigma c}$. I'm trying to see if I can use that somewhere in the integral but I'm a bit stuck on how. Any thoughts on how to proceed from here would be appreciated.