For real square matrices $A_0,A_1$ with $(A^{T}_0+A_0)+(A^{T}_{1}+A_1)\leq{-\gamma{I}}$, is it possible to prove the following inequality? For real square matrices $A_0$ and $A_{1}$ with $A^{T}_0+A_0+A^{T}_{1}+A_1\leq{-\gamma{I}}$, where $\leq$ denotes the negative semi-definiteness of the left argument with respect to the right, $\gamma>0$ and $I$ is the identity matrix of suitable dimension.
Is it possible to show that 
$x^T\Big(e^{A^{T}_{1}}*e^{A^{T}_{0}}*e^{A_{0}}*e^{A_{1}}\Big)x\leq\|x\|^2$ for a non-zero vector $x$? I know if $A_0$ and $A_1$ are normal matrices and $A_0, A_1$ are commutative to each other, the mentioned claim holds immediately. But for the case when these matrices are not normal and do not commute, does the claim hold?
My approach:
$x^T\Big(e^{A^{T}_{1}}*e^{A^{T}_{0}}*e^{A_{0}}*e^{A_{1}}\Big)x\leq\|x\|^2\|e^{A_0}*e^{A_1}\|^2\leq\|x\|^2\|e^{A_0}*e^{A_1}\|_F^2=\|x\|^2~\|e^{A_0}\|_F^2\|e^{A_1}\|_F^2=\|x\|^2\text{Tr}\left(e^{A^T_0}e^{A_0}\right)\text{Tr}\left(e^{A^T_1}e^{A_1}\right)\leq\|x\|^2\text{Tr}\left(e^{A^T_0+A_0}\right)\text{Tr}\left(e^{A^T_1+A_1}\right)$ by virtue of equation (1.3) of 
 https://epubs.siam.org/doi/pdf/10.1137/0609012. 
But I do not see a way forward. Any hints or suggestions are greatly appreciated.
 A: Let $A$ and $B$ be such that $A$ and $B$ are symmetric and $2A \leq -2B$ but $e^{2A} \leq e^{-2B}$ does not hold. For example, following this post one can set
$$A=\begin{pmatrix}0.5 & 0 \\ 0 & 1\end{pmatrix},~B=\begin{pmatrix}-1-\epsilon & -0.5 \\-0.5 & -1.5\end{pmatrix},$$
where $\epsilon>0$ is small enough. Also let $A_0=A_0^T=0.5A$ and $A_1=A_1^T=0.5 B$. Then we have $A_0+A_0^T+A_1+A_1^T=A+B\leq 0$. The desired inequality is equivalent to the claim that $\|e^{A_1^T}e^{A_0^T}e^{A_0}e^{A_1}\|_{op}\leq 1$. We prove this is false for our choice of $A_0$ and $A_1$. Since $A_0$ and $A_1$ are symmetric (hence normal):
$$\| e^{A_1^T}e^{A_0^T}e^{A_0}e^{A_1}\|_{op}=\|e^{A_0^T}e^{A_0}e^{A_1}e^{A_1^T}\|_{op}=\|e^{A}e^{B}\|_{op}.$$
So suppose on the contrary that $\|e^Ae^B \|_{op} \leq 1$, and so $\|e^Ae^Bx\|\leq \|x\|$ for all $x$. Therefore, $\|e^A(e^Bx)\|\leq \|e^{-B}(e^Bx)\|$ for all $x$. Since $e^B$ is invertible, it follows that $\|e^Ax \|\leq \|e^{-B}x\|$ for all $x$. Equivalently, $\langle e^{2A}x,x\rangle \leq \langle e^{-2B}x,x \rangle$ for all $x$ and so $\langle (e^{2A}-e^{-2B})x,x \rangle \leq 0$. But then $e^{2A}-e^{-2B}\leq 0$, which is a contradiction. 
