An inequality for expected value of normally distributed variables

The problem posed above is a semi-well-known open problem that I believe is equivalent to the real polarization conjecture. A more general version of the question posed above is offered as Conjecture 4 in this paper of Wenbo Li, who considers arbitrary positive powers of the type $|X_i|^{p_i}$.

For the case $k=1$, a proof is known. Have a look at Theorem 2.1 in the linked paper for a proof of $k=1$ case (thanks to Iosif for alerting me that my answer was only about the $k=1$ case). In particular, the variables have to be jointly Gaussian with mean zero. Moreover, equality holds if and only if they are independent or at least one of them is a.s. zero.

Some partial counterexamples:

Let us assume that "with normal distribution" means "with joint normal distribution". Even then, in general the inequality will not hold. Indeed, \begin{equation} E(Z-1)^2(Z+1)^2<E(Z-1)^2\;E(Z+1)^2, \end{equation} where $Z\sim N(0,1)$.

The inequality may hold assuming also that the $X_i$'s are zero-mean.

However, the stronger inequality \begin{equation} E \prod_{i\in S\cup T}X_i^{2k}\ge E \prod_{i\in S}X_i^{2k}\ E \prod_{i\in T}X_i^{2k} \end{equation} for finite disjoint sets $S$ and $T$ will not hold in general (or even when the set $T$ is a singleton) -- even if the $X_i$'s are zero-mean jointly normal. E.g., let $X_1=Z_1$, $X_2=Z_1-Z_2$, $X_3=\frac12 Z_1+Z_2$, where the $Z_i$'s are independent $N(0,1)$. Then \begin{equation} E X_1^2 X_2^2 X_3^2=\frac{15}4<4\times\frac54=E X_1^2 X_2^2 \ E X_3^2. \end{equation}