Convolution integral inequality, with a special case given by Young's inequality For every $n\in\mathbb N$ and $t>0$, let $\gamma^n_t$ be the $n$-dimensional Gaussian density
$$\gamma^n_t(x):=\frac{\exp(-\|x\|^2/2t)}{(2\pi t)^{n/2}},\qquad x\in\mathbb R^n.$$
For simplicity let $\gamma^n:=\gamma^n_1$.

Numerical simulations (and exact computations in the case $n=1$) I've done suggest that the following is true. For any $n\times n$ matrices $A$ and $B$, and $t>0$, it holds that
\begin{align}\tag{1}
\int\left(\int\gamma_t^n(x-u)\gamma^n(Au) ~du\right)\left(\int\gamma_t^n(x-u)\gamma^n(Bu) ~du\right)~dx\leq\int\gamma^n(Ax)\gamma^n(Bx)~dx
\end{align}
(assuming everything is integrable). In other words, if we approximate the functions $x\mapsto\gamma^n(Ax)$ and $x\mapsto\gamma^n(Bx)$ by convolving each of them with $\gamma^n_t$, and then take the integral of their product, this is smaller or equal to the integral of $\gamma^n(Ax)\gamma^n(Bx)$.

In the very special case where $A=B$, this is a direct consequence of Young's convolution inequality:
\begin{align}
\int\left(\int\gamma_t^n(x-u)\gamma^n(Au) ~du\right)^2~dx
&\leq\left(\int\gamma^n_t(x)~dx\right)^2\left(\int\big(\gamma^n(Ax)\big)^2~dx\right)\\
&=\int\big(\gamma^n(Ax)\big)^2~dx.
\end{align}

Question. Is there an easy way to see (if true) why $(1)$ holds in general when $A\neq B$?


Attempts on my part (1). I've tried to see if the general case can somehow be reduced to the $A=B$ case, but this doesn't seem to work nicely at all. For example in $1$-D:
$$\int \gamma^1(a x)\gamma^1(b x)~dx=\frac{1}{\sqrt{2 \pi (a^2+b^2)}}=\int \big(\gamma^1(k x)\big)^2~dx$$
with $k=\frac{\sqrt{s^2+t^2}}{\sqrt{2}}$, and for $t>0$
\begin{align}
&\int\left(\int\gamma_t^1(x-u)\gamma^1(au) ~du\right)\left(\int\gamma_t^1(x-u)\gamma^1(bu) ~du\right)~dx\\
&=\frac{1}{\sqrt{2 \pi(2 t a^2 b^2+a^2+b^2)}}\\
&=\int\left(\int\gamma_t^1(x-u)\gamma^1(k_tu) ~du\right)^2~dx
\end{align}
with $k_t=\sqrt{\frac{1}{t^2}+\frac{2 a^2}{t}+\frac{2 b^2}{t}+4 a^2 b^2}-\frac{1}{t}$. We can immediately see that these computations prove my claim for $a\neq b$ when $n=1$, but my hope here was to somehow combine Young's inequality with
$$\int \big(\gamma^1(k_t x)\big)^2~dx\leq\int \big(\gamma^1(k x)\big)^2~dx,$$
to get a general proof for $n\geq2$, but this becomes extremely unwieldy in higher dimensions.
(2). I've tried to look at generalizations of Young's convolution inequality, and found the Brascamp-Lieb inequality, but it doesn't look like this helps in this case at all.
 A: It isn't too hard to make your 1-D argument rigorous in higher-dimensions.  
Note that each Hermitian, positive-semidefinite matrix $\renewcommand{\d}{\,\mathrm{d}}$ $\mathcal{A}$ has a well-defined square root as follows: because $\mathcal{A}$ is Hermitian, it is diagonalizable.   Let the eigenvalues and eigenspaces be $\{(\lambda_k,V_k)\}$; by definition, we may identify $\mathbb{R}^n\cong\bigoplus_k{V_k}$.   By positive-semidefiniteness, each $\lambda_l\geq0$.   So we define $$\sqrt{\mathcal{A}}(\bigoplus{x_k})=\sum_k{\sqrt{\lambda_k}x_k}$$  Note that we furthermore have $[\sqrt{\mathcal{A}},\mathcal{A}]=0$.  
Let $\alpha=I_n+tA^{\intercal}A$.  $\alpha$ is positive-semidefinite Hermitian, so it has a square root; we claim that $$\gamma_t^n*(\gamma^n\circ A)=\frac{\gamma^n\circ A\sqrt{\alpha}^{-1}}{\det{\left(\sqrt{\alpha}\right)}}$$  Similarly, let $\beta=I_n+tB^{\intercal}B$; then $$\gamma_t^n*(\gamma^n\circ B)=\frac{\gamma^n\circ B\sqrt{\beta}^{-1}}{\det{(\sqrt{\beta})}}$$  
To see this, calculate \begin{align*}
(\gamma_t^n*(\gamma^n\circ A))(u)&=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^n}{(2\pi t)^{-\frac{n}{2}}\exp{\left(-\frac{|x-u|^2}{2t}-\frac{|Ax|^2}{2}\right)}\d^nx} \\
&=(2\pi)^{-\frac{n}{2}}e^{-\frac{|u|^2}{2t}}\int_{\mathbb{R}^n}{(2\pi t)^{-\frac{n}{2}}\exp{\left(\frac{x^{\intercal}(I_n+tA^{\intercal}A)x-2u^{\intercal}x}{2t}\right)}\d^nx}
\end{align*}  Completing the square, we have \begin{align*}
(\gamma_t^n*(\gamma^n\circ A))(u)&=(2\pi)^{-\frac{n}{2}}e^{\frac{|\sqrt{\alpha}^{-1}u|^2-|u|^2}{2t}}\int_{\mathbb{R}^n}{\gamma_t^n\left(\sqrt{\alpha}x-\sqrt{\alpha}^{-1}u\right)\d^nx} \\
&=\frac{(2\pi)^{-\frac{n}{2}}e^{\frac{|\sqrt{\alpha}^{-1}u|^2-|u|^2}{2t}}}{\det{\left(\sqrt{\alpha}\right)}}
\end{align*} by change of variables, noting that $\|\gamma_t^n\|_1=1$.  Now, \begin{align*}
\frac{|\sqrt{I_n+tA^{\intercal}A}^{-1}u|^2-|u|^2}{2t}&=\frac{u^{\intercal}\left(\sqrt{I_n+tA^{\intercal}A}^{-1}\right)(I_n-(I_n+tA^{\intercal}A))\left(\sqrt{I_n+tA^{\intercal}A}^{-1}\right)u}{2t} \\
&=\frac{1}{2}|A\sqrt{I_n+tA^{\intercal}A}^{-1}u|^2
\end{align*}  Substituting, we obtain the claimed result.  
Now, \begin{align*}
\int_{\mathbb{R}^n}{\gamma^n(Ax)\gamma^n(Bx)\d^nx}&=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^n}{(2\pi)^{-\frac{n}{2}}e^{-\frac{1}{2}x^{\intercal}(A^{\intercal}A+B^{\intercal}B)x}\d^nx} \\
&=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^n}{\gamma^n\left(\sqrt{A^{\intercal}A+B^{\intercal}B}x\right)\d^nx} \\
&=\frac{(2\pi)^{-\frac{n}{2}}}{\det{(\sqrt{A^{\intercal}A+B^{\intercal}B})}}
\end{align*}  Similarly, the left-hand side of (1) is $$\mathrm{LHS}=\frac{(2\pi)^{-\frac{n}{2}}}{\det{(\sqrt{\alpha})}\det{(\sqrt{\beta})}\det{(\sqrt{\sqrt{\alpha}^{-1}A^{\intercal}A\sqrt{\alpha}^{-1}+\sqrt{\beta}^{-1}B^{\intercal}B\sqrt{\beta}^{-1}})}}$$  
Now, by construction, $\det{(\sqrt{M})}=\sqrt{\det{(M)}}$, where defined.   Furthermore, $[\sqrt{\alpha}^{-1},\alpha]=0$ and $[\alpha,A^{\intercal}A]=[I_n+A^{\intercal}A,A^{\intercal}A]=0$.   So $[\sqrt{\alpha}{^-1},A^{\intercal}A]=0$, and similarly $[\sqrt{\beta}^{-1},B^{\intercal}B]=0$.   So we may simplify the left-hand side as \begin{align*}
\mathrm{LHS}&=(2\pi)^{-\frac{n}{2}}\det{(\alpha(\alpha^{-1}A^{\intercal}A+B^{\intercal}B\beta^{-1})\beta)}^{-\frac{1}{2}}\\
&=(2\pi)^{-\frac{n}{2}}\det{(A^{\intercal}A\beta+\alpha B^{\intercal}B)}^{-\frac{1}{2}}
\end{align*}  
Thus, (1) is equivalent to $$(2\pi)^{-\frac{n}{2}}\det{(A^{\intercal}A\beta+\alpha B^{\intercal}B)}^{-\frac{1}{2}}\leq\frac{(2\pi)^{-\frac{n}{2}}}{\sqrt{\det{(A^{\intercal}A+B^{\intercal}B)}}}$$ or, clearing constants and denominators, $$\det{(A^{\intercal}A+B^{\intercal}B)}\leq\det{(A^{\intercal}A\beta+\alpha B^{\intercal}B)}$$  Well, all the matrices involved are Hermitian, positive semi-definite, so by Minkowski's determinant inequality, \begin{align*}
\sqrt[n]{\det{(A^{\intercal}A\beta+\alpha B^{\intercal}B)}}&=\sqrt[n]{\det{(A^{\intercal}A+B^{\intercal}B+A^{\intercal}AB^{\intercal}B+B^{\intercal}BA^{\intercal}A)}} \\
&\geq\sqrt[n]{\det{(A^{\intercal}A+B^{\intercal}B)}}+\sqrt[n]{\det{(A^{\intercal}AB^{\intercal}B+B^{\intercal}BA^{\intercal}A)}} \\
&\geq\sqrt[n]{\det{(A^{\intercal}A+B^{\intercal}B)}}
\end{align*}  Clearing $n$th roots, we obtain the desired result.  
