# Comparing two multivariate normal distribution

Let $$\mathbf{Z}\sim N(\boldsymbol{\mu},\mathrm{\Sigma})$$, where $$$$\label{Eq.Mean} \boldsymbol{\mu}^{\rm T} = \delta[-\sqrt{\frac{x_1x_2}{x_1+x_2}},\frac{-1}2\sqrt{\frac{x_1x_3}{x_1+x_3}},\frac12\sqrt{\frac{x_2x_3}{x_2+x_3}}]$$$$ and

$$$$\mathrm{\Sigma} = \begin{bmatrix}1 & \frac{x_2x_3}{(1-x_2)(1-x_3)} & \frac{-x_1x_3}{(1-x_1)(1-x_3)}\\ \frac{x_2x_3}{(1-x_2)(1-x_3)} & 1 & \frac{x_1x_2}{(1-x_1)(1-x_2)}\\ \frac{-x_1x_3}{(1-x_1)(1-x_3)} & \frac{x_1x_2}{(1-x_1)(1-x_2)}& 1 \end{bmatrix},$$$$ $$0\leq x_i\leq 1$$, $$x_1+x_2+x_3 = 1$$ and $$\delta>0$$. Further suppose for some $$a>1$$ $$$$\psi_x = 1 - \int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}\phi(\boldsymbol{\mu};\mathrm{\Sigma})d\boldsymbol{z}.$$$$

When $$x_1 = x_2 = x_3 = 1/3$$, then $$$$\boldsymbol{\mu}^{\rm T} = \frac{\delta}{2\sqrt{6}}[-2,-1,1]$$$$ and

$$$$\mathrm{\Sigma} = \begin{bmatrix}1 & \frac{1}{2} & \frac{-1}{2}\\ \frac{1}{2} & 1 & \frac{1}{2}\\ \frac{-1}{2} & \frac{1}{2} & 1 \end{bmatrix}.$$$$ Now let $$y_1=\min(x_1,x_2,x_3)$$, $$y_2=\text{min}\{(x_1,x_2,x_3)\setminus y_1\}$$ and $$y_3=\max(x_1,x_2,x_3)$$. Let $$\psi_y$$ be the result of replacing $$\boldsymbol{x} = (x_1,x_2,x_3)$$ by $$\boldsymbol{y} = (y_1, y_2, y_3)$$ in the general expressions of $$\boldsymbol{\mu}$$, $$\mathrm{\Sigma}$$ and $$\psi$$.

I am able to show numerically that $$\psi_{\boldsymbol{x}=(1/3,1/3,1/3)}>\psi_y$$. Can it be proved analytically?