The inequality is a special case of the following claim.

Claim:
If $X = (X_1, \dotsc, X_d) \sim N(\mu, \Sigma)$ is an $\mathbb{R}^d$-valued normal random variable with invertible covariance matrix $\Sigma$, $L : \mathbb{R}^d \to \mathbb{R}$ is linear, and $K \subseteq \mathbb{R}^d$ is convex and has a nonempty interior, then $\operatorname{Var}(L(X) \mid X \in K) \le \operatorname{Var}(L(X))$, with equality if and only if $L(X)$ and $I(X \in K)$ are independent.
(Here $I(X \in K)$ is the indicator function of the event $X \in K$.)

Proof:
We may assume that $L \neq 0$.
After a linear change of coordinates, we may assume that $X$ is standard normal and $L(X) = X_1$.
Then the PDF of $X_1$ is $\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}}$, and the PDF of $Z = (X_1 \mid X \in K)$ is $\varphi(x) = C e^{-\frac{x^2}{2}} P(X' \in K_x)$, where $X' = (X_2, \dotsc, X_d)$,
$$
K_x = \{(x_2, \dotsc, x_d) \in \mathbb{R}^{d-1}; \, (x, x_2, \dotsc, x_d) \in K\}
$$
and $C > 0$.
The function $\psi \colon \mathbb{R} \to [0,1]$, $x \mapsto P(X' \in K_x)$ is log-concave:
if $x, y \in \mathbb{R}$ and $\lambda \in (0,1)$, then
$$
\psi((1-\lambda) x + \lambda y) \ge P(X' \in (1-\lambda) K_x + \lambda K_y) \ge \psi(x)^{1-\lambda} \psi(y)^{\lambda},
$$
because normal distributions are log-concave (see https://en.wikipedia.org/wiki/Logarithmically_concave_measure).
Note that $X_1$ and $I(X \in K)$ are independent if and only if $\psi$ is a constant function.
The Claim follows from the following lemma.

Lemma: If $Z$ is a real random variable with PDF $\varphi(x)$ such that $e^{\frac{x^2}{2}} \varphi(x)$ is log-concave, then $\operatorname{Var}(Z) \le 1$, with equality if and only if $Z \sim N(E[Z],1)$.

Proof:
We may add a constant to $Z$, so we may assume that $\varphi$ is maximal at $0$.
Then $\varphi$ is decreasing in $\mathbb{R}_{\ge 0}$, increasing in $\mathbb{R}_{\le 0}$, and $\varphi(x) \le \varphi(0) e^{-\frac{x^2}{2}}$ for every $x \in \mathbb{R}$.
We have
\begin{align}
& 1 - \operatorname{Var}(Z) \ge \int_{\mathbb{R}} (1-x^2) \varphi(x) \, dx \\[6pt]
= {} & \int_0^\infty (1-x^2) \varphi(x) \, dx + \int_0^\infty (1-x^2) \varphi(-x) \, dx.
\end{align}
We prove that $\int_0^\infty (1-x^2) \varphi(x) \, dx \ge 0$.
This is clear if $\varphi(1) = 0$, so let $\varphi(1) > 0$.
There is a $c \ge 0$ such that $\tilde{\varphi}(1) = \varphi(1)$, where $\tilde{\varphi}(x) = \varphi(0) e^{-c x - \frac{x^2}{2}}$.
Then $\varphi(x) \ge \tilde{\varphi}(x)$ for $x \in [0,1]$, and $\varphi(x) \le \tilde{\varphi}(x)$ for $x \in [1,\infty)$, so $\int_0^\infty (1-x^2) \varphi(x) \, dx \ge \int_0^\infty (1-x^2) \tilde{\varphi}(x) \, dx = \varphi(0) \int_0^\infty (1-x^2) e^{-c x - \frac{x^2}{2}} \, dx$.
We can do the same for $\int_0^\infty (1-x^2) \varphi(-x) \, dx$.
So all we need now is that
$$
\int_0^\infty (1-x^2) e^{-c x - \frac{x^2}{2}} \, dx > 0
$$
for $c > 0$.
We have $\int_0^\infty e^{-c x - \frac{x^2}{2}} \, dx = \frac{\sqrt{\pi}}{\sqrt{2}} e^{\frac{c^2}{2}} \operatorname{erfc}(\frac{c}{\sqrt{2}})$, and differentiating twice in $c$ we get $\int_0^\infty x^2 e^{-c x - \frac{x^2}{2}} \, dx$.
In the end, the inequality boils down to $c e^{\frac{c^2}{2}} \operatorname{erfc}(\frac{c}{\sqrt{2}}) < \frac{\sqrt{2}}{\sqrt{\pi}}$.
Equivalently, $\frac{\sqrt{\pi}}{2} \operatorname{erfc}(x) < x^{-1} e^{-x^2}$ for $x > 0$.
This follows from $(\frac{\sqrt{\pi}}{2} \operatorname{erfc}(x) - x^{-1} e^{-x^2})' = (1+x^{-2}) e^{-x^2} > 0$, since the limits at $x \to \infty$ are $0$.

isindependent of $X_1$ and $X_3$, but it feels like there is a weird dependence between them introduced by the inequalities? For example, cases where $X_1 >> b$ and $X_2 >> X_1$ are very unlikely under both inequalities, but become more likely under just the one. It's possible I am making a logical error though. $\endgroup$`X_2\gg X_1`

. MathJax is as good at some things as LaTeX is, and LaTeX is very very good. Just enter "LaTeX symbols" into a search engine and you can find things. $\endgroup$1more comment