I am trying to prove a conditional gaussian comparison inequality. Specifically, I am looking to prove that for two $n$-dimensional Gaussian variables $X_1,\cdots,X_n$ and $Y_1,\cdots,Y_n$, assume that $\mathbb{E}[X_i]=\mathbb{E}[Y_i]=0$, $\mathbb{E}[X_i^2]=\mathbb{E}[Y_i^2]$, $\mathbb{E}[X_iX_j]\leq \mathbb{E}[Y_iY_j]$, we have for all $u_1,\cdots, u_{n-1} < 0$: $$\mathbb{P}\left[\bigcap_{i=1}^{n-1} \{X_i \geq u_i \} \mid X_n=0 \right]\leq \mathbb{P}\left[\bigcap_{i=1}^{n-1} \{Y_i \geq u_i \}\mid Y_n=0 \right]$$ Intuitively this makes sense because $Y_i$ are more correlated with each other and thus should have a lower tail probability of being all far away from zero, but I have having a hard time proving it rigorously.
I know Kahane's inequality and thought it can get me there, but the fact that I am conditioning on $X_n/Y_n=0$ makes a direct application difficult, so I was wondering if I am overlooking an easier approach.
This is false if you include $u_i=0$, here is a counterexample.
Take $n=2$, let $X_1,Y_1,Y_2$ be independent standard Gaussians, and $X_2=-X_1$. All have mean zero and unit variance, and $\mathbb{E}[X_1X_2]=-1<0=\mathbb{E}[Y_1Y_2]$. Moreover, for $u_1=0$, $$ \mathbb{P}\left[X_1\geq 0|X_2=0\right]=1 >\frac{1}{2}=\mathbb{P}\left[Y_1\geq 0|Y_2=0\right] $$
Edit: For $u_i<0$, the same counterexample works: $$ \mathbb{P}\left[X_1\geq u_1|X_2=0\right] \geq\mathbb{P}\left[X_1= 0|X_2=0\right] =1 >\mathbb{P}\left[Y_1\geq u_1\right]=\mathbb{P}\left[Y_1\geq u_1|Y_2=0\right] $$